Otdfbt

Can an old DSLR be upgraded to match modern smartphone image quality

Is having a hidden directory under /etc safe?

How should I push back against my job assigning "homework"?

Thousands and thousands of words

Is this light switch installation safe and legal?

How crucial is a waifu game storyline?

How do I subvert the tropes of a train heist?

Uncommanded roll at high speed

Different PCB color ( is it different material? )

How to capture more stars?

chmod would set file permission to 000 no matter what permission i try to set

Fastest way to perform complex search on pandas dataframe

Asking bank to reduce APR instead of increasing credit limit

Looking after a wayward brother in mother's will

Can a non-EU citizen travel within the Schengen area without identity documents?

What was this black-and-white film set in the Arctic or Antarctic where the monster/alien gets fried in the end?

Can a rogue effectively triple their speed by combining Dash and Ready?

Did airlines fly their aircraft slower in response to oil prices in the 1970s?

Strange math syntax in old basic listing

Are UK pensions taxed twice?

When a current flow in an inductor is interrupted, what limits the voltage rise?

Beginner's snake game using PyGame

Are there regional foods in Westeros?

Do creatures all have the same statistics upon being reanimated via the Animate Dead spell?

What formula to chose a nonlinear formula?

Numerical integration of a dataset with method other than the trapezoidal ruleHow To Create a Non-Linear Output from a Linear Input?What is the simplest formula for activation / smooth step function?Process For Building a Function?“Proportional to” - but nonlinear.Range of nonlinear functionFunctional equation - nonlinearAsymptotic functions with derivatives that are $1/2^x$Validating a function over a domain in a computer program.(nonlinear) Regression, but: What type of function is this?What is the most simple formula to achieve this pattern?

1

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  So you want a function $f(x)$ such that $f(0) = 0$ , $f(128)=150$ and $f(255) = 255$?
  $endgroup$
  – NoChance
  May 15 at 21:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please refer to this MathJax tutorial.
  $endgroup$
  – Pantelis Sopasakis
  May 15 at 21:17
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may use a logistic function.
  $endgroup$
  – minori minus
  May 15 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can interpolate and get a parabolic function, I don't get what you mean by "opposite as a parabolic function".
  $endgroup$
  – Crostul
  May 15 at 21:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See wolframalpha.com/input/… . Wolfram Alpha describes an arc of a parabola.
  $endgroup$
  – Crostul
  May 15 at 21:26
  

  
  
  
  

 | 
show 4 more comments

1

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  So you want a function $f(x)$ such that $f(0) = 0$ , $f(128)=150$ and $f(255) = 255$?
  $endgroup$
  – NoChance
  May 15 at 21:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please refer to this MathJax tutorial.
  $endgroup$
  – Pantelis Sopasakis
  May 15 at 21:17
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may use a logistic function.
  $endgroup$
  – minori minus
  May 15 at 21:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can interpolate and get a parabolic function, I don't get what you mean by "opposite as a parabolic function".
  $endgroup$
  – Crostul
  May 15 at 21:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See wolframalpha.com/input/… . Wolfram Alpha describes an arc of a parabola.
  $endgroup$
  – Crostul
  May 15 at 21:26
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

$endgroup$

share|cite|improve this question

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

share|cite|improve this question

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

edited May 15 at 21:28

Pantelis Sopasakis

2,7791141

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

asked May 15 at 21:13

Michel KeijzersMichel Keijzers

1469

3 Answers
3

active

oldest

votes

6

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered May 15 at 21:31

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

6

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered May 15 at 21:59

zwimzwim

13.5k834

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  May 15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichelKeijzers In that case, I'd wager there already exists a library function for precisely that purpose
  $endgroup$
  – Hagen von Eitzen
  May 16 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Desmos seems like an excellent tool. Thanks for the tip!
  $endgroup$
  – Eric Duminil
  May 16 at 6:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HagenvonEitzen Probably yes, but for this so simple function I can program the one line formula.
  $endgroup$
  – Michel Keijzers
  May 16 at 11:09
  

  
  
  
  

add a comment | 

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited May 15 at 21:41

answered May 15 at 21:37

NoChanceNoChance

3,97521321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  May 15 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  May 15 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:50
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3227585%2fwhat-formula-to-chose-a-nonlinear-formula%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered May 15 at 21:31

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

6

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered May 15 at 21:31

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered May 15 at 21:31

CrostulCrostul

28.4k22352

$endgroup$

share|cite|improve this answer

answered May 15 at 21:31

CrostulCrostul

28.4k22352

answered May 15 at 21:31

CrostulCrostul

28.4k22352

answered May 15 at 21:31

CrostulCrostul

28.4k22352

6

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered May 15 at 21:59

zwimzwim

13.5k834

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  May 15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichelKeijzers In that case, I'd wager there already exists a library function for precisely that purpose
  $endgroup$
  – Hagen von Eitzen
  May 16 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Desmos seems like an excellent tool. Thanks for the tip!
  $endgroup$
  – Eric Duminil
  May 16 at 6:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HagenvonEitzen Probably yes, but for this so simple function I can program the one line formula.
  $endgroup$
  – Michel Keijzers
  May 16 at 11:09
  

  
  
  
  

add a comment | 

6

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered May 15 at 21:59

zwimzwim

13.5k834

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  May 15 at 22:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichelKeijzers In that case, I'd wager there already exists a library function for precisely that purpose
  $endgroup$
  – Hagen von Eitzen
  May 16 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Desmos seems like an excellent tool. Thanks for the tip!
  $endgroup$
  – Eric Duminil
  May 16 at 6:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HagenvonEitzen Probably yes, but for this so simple function I can program the one line formula.
  $endgroup$
  – Michel Keijzers
  May 16 at 11:09
  

  
  
  
  

add a comment | 

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered May 15 at 21:59

zwimzwim

13.5k834

$endgroup$

share|cite|improve this answer

answered May 15 at 21:59

zwimzwim

13.5k834

answered May 15 at 21:59

zwimzwim

13.5k834

answered May 15 at 21:59

zwimzwim

13.5k834

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited May 15 at 21:41

answered May 15 at 21:37

NoChanceNoChance

3,97521321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  May 15 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  May 15 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:50
  

  
  
  
  

add a comment | 

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited May 15 at 21:41

answered May 15 at 21:37

NoChanceNoChance

3,97521321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  May 15 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  May 15 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  May 15 at 21:50
  

  
  
  
  

add a comment | 

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited May 15 at 21:41

answered May 15 at 21:37

NoChanceNoChance

3,97521321

$endgroup$

share|cite|improve this answer

edited May 15 at 21:41

answered May 15 at 21:37

NoChanceNoChance

3,97521321

answered May 15 at 21:37

NoChanceNoChance

3,97521321

answered May 15 at 21:37

NoChanceNoChance

3,97521321