162
\$\begingroup\$

A truth-machine (credits goes to this guy for coming up with it) is a very simple program designed to demonstrate the I/O and control flow of a language. Here's what a truth-machine does:

  • Gets a number (either 0 or 1) from STDIN.

  • If that number is 0, print out 0 and terminate.

  • If that number is 1, print out 1 forever.

Challenge

Write a truth-machine as described above in your language of choice. The truth-machine must be a full program that follows these rules:

  • take input from STDIN or an acceptable alternative
    • If your language cannot take input from STDIN, it may take input from a hardcoded variable or suitable equivalent in the program
  • must output to STDOUT or an acceptable alternative
    • If your language is incapable of outputting the characters 0 or 1, byte or unary I/O is acceptable.
  • when the input is 1, it must continually print 1s and only stop if the program is killed or runs out of memory
  • the output must only be either a 0 followed by either one or no newline or space, or infinite 1s with each 1 followed by either one or no newline or space. No other output can be generated, except constant output of your language's interpreter that cannot be suppressed (such as a greeting, ANSI color codes or indentation). Your usage of newlines or spaces must be consistent: for example, if you choose to output 1 with a newline after it all 1s must have a newline after them.

  • if and only if your language cannot possibly terminate on an input of 0 it is acceptable for the code to enter an infinite loop in which nothing is outputted.

Since this is a catalog, languages created after this challenge are allowed to compete. Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language. Other than that, all the standard rules of must be obeyed. Submissions in most languages will be scored in bytes in an appropriate preexisting encoding (usually UTF-8).

Catalog

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 320px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 62732; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>

\$\endgroup\$
18
  • \$\begingroup\$ Can we assume that the program halts when the processor finishes executing the written code, for a machine code entry? \$\endgroup\$
    – lirtosiast
    Nov 3 '15 at 16:58
  • 3
    \$\begingroup\$ Assuming any behaviour is fine for all invalid inputs? \$\endgroup\$
    – Cruncher
    Nov 3 '15 at 17:33
  • 3
    \$\begingroup\$ @Cruncher Yes, the only inputs you should expect to get are 0 and 1. \$\endgroup\$ Nov 3 '15 at 17:38
  • 4
    \$\begingroup\$ Catalog is borked. \$\endgroup\$ Nov 6 '15 at 15:18
  • 2
    \$\begingroup\$ Catalog appears to consider Bf and bf to be different languages. \$\endgroup\$ Nov 10 '15 at 1:13

445 Answers 445

1
11 12 13 14
15
0
\$\begingroup\$

sed 4.2.2, 8 bytes

Uses the -n flag to suppress printing without an explicit p command.

:
p
/1/b

Try it online!

Commented:

#create an unnamed label
:
#print pattern space
p
#If pattern space contains 1, branch to beginning
/1/b
\$\endgroup\$
0
\$\begingroup\$

dotcomma, 8 bytes

[,].[.,]

Try it online!

Dotcomma is a language I recently invented, designed to do things in an interesting way with as few instructions as possible. In order to accomplish this, there are two operators (predictably . and ,) which can do entirely different things depending on context. This answer's explanation will be very introductory. To see some more complicated code, check out the examples in the page linked in the title.

This program is actually a pretty neat showcase of some of the more interesting features of dotcomma. You'll notice it's divides into two main parts, [,] and [.,], separated by a .. These are called blocks. Every block and operator has a return value.

The way dotcomma manages to do so many different things with each operator is by changing what they do based on what precedes and follows them. Take the , in the first block. Because it is at the start of a block, its return value will be a value taken from input. Because it is at the end of a block, it will also output this value and use it as the block's return value. Surprisingly compact for a language with four valid characters!

As for the first ., because it is preceded by a block, it will take the block's return value and use it as its own. It is followed by another block, so it will do one two things: if its return value is 0, the block will be skipped. Thus, only 0 will be outputted and the program will terminate.

However, if its return value is 1, the block will be looped. Within the block are two operators: .,. The . will have a return value of 1, as it follows the beginning of a block. The , will behave similarly to the first one, but instead of taking the number to be outputted from input, it will take it from the . (which will always be 1). The loop will run forever because the , will always set the looped block's return value to 1.

Hopefully this explanation isn't too confusing!

\$\endgroup\$
0
\$\begingroup\$

Fugue, 120 bytes

00000000: 4d54 6864 0000 0006 0001 0002 0001 4d54  MThd..........MT
00000010: 726b 0000 002b 0090 4040 0290 3e40 0190  rk...+..@@..>@..
00000020: 3a40 0090 4040 0190 4740 0190 4340 0090  :@..@@..G@..C@..
00000030: 4440 0190 3e40 0190 3740 0190 3740 00ff  D@..>@..7@..7@..
00000040: 2f4d 5472 6b00 0000 2f00 9040 4001 9037  /MTrk.../..@@..7
00000050: 4001 903f 4001 9040 4002 903c 4000 9044  @..?@..@@..<@..D
00000060: 4001 903e 4002 9045 4001 9046 4001 904e  @..>@..E@..F@..N
00000070: 4001 9047 4000 ff2f                      @..G@../

Since the Fugue compiler does not support voice wrapping, a direct translation of Martin Ender's Prelude answer turns out to not be the shortest possible solution. This program is equivalent to the Prelude program:

 v6(1-)#
?!^ 8- (^!)

See my Hello World answer for more information on how to use the compiler. Finally, here's a valid MIDI version of the program:

00000000: 4d54 6864 0000 0006 0001 0002 0001 4d54  MThd..........MT
00000010: 726b 0000 002c 0090 4040 0290 3e40 0190  rk...,..@@..>@..
00000020: 3a40 0090 4040 0190 4740 0190 4340 0090  :@..@@..G@..C@..
00000030: 4440 0190 3e40 0190 3740 0190 3740 00ff  D@..>@..7@..7@..
00000040: 2f00 4d54 726b 0000 0030 0090 4040 0190  /.MTrk...0..@@..
00000050: 3740 0190 3f40 0190 4040 0290 3c40 0090  7@..?@..@@..<@..
00000060: 4440 0190 3e40 0290 4540 0190 4640 0190  D@..>@..E@..F@..
00000070: 4e40 0190 4740 00ff 2f00                 N@..G@../.
\$\endgroup\$
0
\$\begingroup\$

BRASCA, 11 bytes

x48S-:n[:n]

Explanation

<implicit input> - Take input from STDIN
x                - Clean up the linefeed from STDIN
 48S-            - Remove 48 from the digit's ascii code
     :n          - Output the number
       [:n]      - While non-zero: Output the number

Language Link

Github Repo

\$\endgroup\$
0
\$\begingroup\$

Python 3, 47 bytes

Not a winner, but I like the method.

set(iter(lambda x=input():(print(x),x)[1],'0'))

Try it online!

\$\endgroup\$
0
\$\begingroup\$

x86_16 machine code - 20 bytes

3E A0 82 00   MOV AL, DS:[82H]
8A D0         MOV DL, AL
          _LOOP:
B4 02         MOV AH, 02H
CD 21         INT 21H
80 FA 31      CMP DL, "1"
74 F7         JZ _LOOP
             
B8 00 4C      MOV AX, 4C00H
CD 21         INT 21H

Tested in DOSBox :

Input 0

enter image description here

Input 1

If input 1 gives infinite loop

enter image description here

\$\endgroup\$
0
\$\begingroup\$

V (vim), 18 13 bytes

xqqp/1
@qq@qx

Try it online!

Somehow, there was no existing vim answer for this. I was happy to oblige.

-5 bytes from Aaron Miller.

Explanation

xqqp/1
x      delete the input char 
 qq      start macro q
   p     paste the input
    /1   find 1 (exits if 0 isn't matched)

@qq@qx    
@q       recursively call macro q
  q    end macro q
   @q  call macro q
     x delete last char (never gets executed in the case of 1)
\$\endgroup\$
5
  • \$\begingroup\$ Does this meet the "If that number is 1, print out 1 forever." requirement? Trying it online doesn't produce any output \$\endgroup\$ Apr 16 at 16:17
  • \$\begingroup\$ vim doesn't show output on tio, sadly. The keystrokes work correctly in the terminal. \$\endgroup\$
    – Razetime
    Apr 16 at 16:18
  • \$\begingroup\$ Maybe I can upload a gif. \$\endgroup\$
    – Razetime
    Apr 16 at 16:18
  • 1
    \$\begingroup\$ @Razetime 13 bytes - /1 will work in this case instead of :s/1/1, and if you put x before the macro instead of using Y, then you can replace the dk with another x. \$\endgroup\$ Apr 16 at 22:23
  • \$\begingroup\$ @AaronMiller Great, added to the answer. \$\endgroup\$
    – Razetime
    Apr 17 at 2:34
0
\$\begingroup\$

Pinecone, 41 bytes

n:"".input;n="1"?(tru@print:1)|(print:0)
\$\endgroup\$
0
\$\begingroup\$

CSASM v2.3.1, 54 bytes

func main:
in ""
conv i32
.lbl a
dup
dup
print
brtrue a
ret
end

Explanation:

func main:
    ; Get the input and convert it to an integer
    in ""
    conv i32

    .lbl a
        ; Duplicate the input twice
        ; Both the "print" and "brtrue" instructions pop a value from the stack
        ;   so this is necessary to keep a 1 on the stack when looping
        dup
        dup
        print

        ; Truthy values are a non-zero integer or a non-null object
        brtrue a
    ret
end
\$\endgroup\$
0
\$\begingroup\$

CASL II, 72 bytes.

Tested here.

A START
 IN B,1
D OUT B,1
 LAD GR1,48
 XOR GR1,B
 JNZ D
 RET
B DS 1
 END

If you are familiar with assembly languages, you would get it well. It uses standard i/o. It uses ASCII-compatible encoding.

\$\endgroup\$
2
  • \$\begingroup\$ It's stupid that START requires a label. \$\endgroup\$ May 5 at 5:20
  • \$\begingroup\$ Also it's buggy because 1 in IN B,1 and OUT B,1 is an address rather than a literal; maybe buffer overflow may occur when assembled. However, wherever the assembled code starts from, it seemed that those two macros are assembled using LAD command: actually not problem, at least on this emulator. \$\endgroup\$ May 6 at 7:09
0
\$\begingroup\$

Pxem, 0 bytes (content) + 13 bytes (filename).

Filename is:

._.c.n.w1.o.a

Try it online!

As verbose pseudocode

push numeircal integer input
dup
unless empty: printf "%d", $(pop)
while one of
  1. empty
  2. $(pop) != 0
do
  push "1", a character
  unless empty: printf "%c", $(pop)
done
\$\endgroup\$
0
\$\begingroup\$

Duocentehexaquinquagesimal, 2 bytes

Try it online! Takes input as a character, and outputs in unary using characters.

\$\endgroup\$
0
\$\begingroup\$

Subleq (8-bit), 15 bytes

-1 15   3
15 16   6
15 -1 -48
16  8  -1
15  8   6

Explanation

  • 0: -1 15 3 Input character to 15:, goto 3
  • 3: 15 16 6 16: = 16: - 15: (16: = -15), goto 6
  • 6: 15 -1 -48 Output 15: (-48 is unused in instruction and used for memory)
  • 9: 16 8 -1 8: = 8: - 16: , if 8: <= 0 exit; exit for all characters "0" and before
  • 12: 15 8 6 8: = 8: - 15: (back to -48), if 8: <= 0 goto 6
\$\endgroup\$
0
\$\begingroup\$

Knight, 13 11 bytes

New method with suggestions from the language's author:

I+0P W1O1O0

Try it online!

# Read a string, coerce to number by adding 0, and check if non-zero
: IF (+ 0 PROMPT) {
    # if it was non-zero, loop printing 1 forever
    : WHILE 1 {
        : OUTPUT 1
    }
    # else output 0 and exit.
    {
        : OUTPUT 0
    }

Old version which I wrote on my own:

;=wP;W+0wO1O0

There are a few ways to do this. Decided this way because of a certain theme.

Ungolfed version:

# Read string into input
; = input PROMPT
# Coerce input to a number by adding 0, and loop while not 0
; WHILE (+ 0 input) {
    # Output 1 and a new line
    : OUTPUT 1
    # this is an infinite loop
}
# Otherwise, output 0 and a new line, and quit
: OUTPUT 0

If we allow excess newlines, this is an arguably superior program:

;=wP;W+0wOwOw

However, this prints two new lines because PROMPT includes the trailing newline. 😔

\$\endgroup\$
0
\$\begingroup\$

Barrel, 10 bytes

This makes use of the fact that the input will be only 0 or 1:

λn?a:#∞n

Explanation:

λ        // input into the accumulator
 n       // print the accumulator
  ? :    // if-else statement (auto-closes to '?a:#∞n::')
   a     // if the accumulator evaluates to truthy (i.e. non-zero)...
     #∞  // ...loop forever...
       n // ...implicitly print the accumulator...
         // ...else do nothing

In other news, I really should make a custom charmap for barrel. With a custom charmap, I could shave off 3 bytes.

\$\endgroup\$
0
\$\begingroup\$

Ocaml, 46 bytes

let a=read_int()in while print_int a;a=1 do 0

urk the functions name and the control flow takes a lot of chars. The trick since there is no do-while is to put the print directly in the condition (sometime ago when I did that in a course lab, the teach called me a "connoisseur of monstrosities" (approx tl))

The 0 in the end causes a warning, so you need to disable werror.

\$\endgroup\$
0
\$\begingroup\$

sed, 13 bytes

:l;s/1/1/p;tl

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Squire, 47 bytes

x=inquire()whence l proclaim(x+" ")if"0"!=x{l:}

Whilst some languages have a goto keyword, Squire useth whence, the opposite. One might call it a comefrom. Squire also containeth a normal loop with whilst, but alas, it is impossible to useth this without enscribing proclaim twice.

If we accompany whence with if, we can create a loop that some call do-whilst.

Ungolfeth:

# Inquire a line from ye standard input
x = inquire()
# Set a whence point
whence loop
# Proclaim x, followed by a space, to ye standard output
proclaim(x + " ")
# If x is not "0"
if "0" != x {
    # Loopeth ad infinitum
    loop:
}

Methinks this language is ridiculous. 😂

\$\endgroup\$
0
\$\begingroup\$

Dis, 49 bytes.

}{*|*^__________!||_^_____________*{___^___****!*

Usage

  • Input from STDIN, as an ASCII number character.
  • Output to STDOUT, in ASCII.

With comments

(
0: }{*|*^
'*'+1: ****!* 

'*'+1: * changed to 48-42or49-42 by |:
   12 10t - 11 20t is 01 20t is 15
   12 11t - 11 20t is 01 21t is 16 

'*'+2: * for * command 

'*'+1: 15or16 for ^ 

15+1: ! reaches when zero
16+1: ||^ reaches when one 

obtw '*'+3: * and '*'+4: * are for | to have the register A have 49 again 

'*'+5: ! for ^ 

After that '*'+4: '1' BUT whatever 

'!'+1: *{___^ repeatedly output one 

* 42 ! 33 

) 

(
0123456789 0123456789 0123456789 0123456789
)
}{*|*^____ ______!||_ ^_________ ____*{___^
(
0123456789 0123456789 0123456789 0123456789
)
___****!*

Try with 0!

Try with 1!

\$\endgroup\$
1
  • \$\begingroup\$ I think I can golf more. \$\endgroup\$ Jun 23 at 10:03
0
\$\begingroup\$

Nim-Lang, 72, 64, 49, 41 (credits to @hyper-neutrino and @Jo King) bytes

var a=readChar(stdin)
while a=='1':echo 1
echo 0

Version with if-statement, with suggested edits:

if readChar(stdin)=='0':echo 0
else:
  while true:echo 1
\$\endgroup\$
4
  • 1
    \$\begingroup\$ I'm not familiar with Nim, but could you remove some whitespace (e.g. after the two colons)? Also, does if'0'==a work instead? Also, is while 1 allowed? That's a trick in Python but I don't know if Nim accepts that. Finally, could you use an if-else instead, or even better, if a is 1, while loop, and then just echo 0 at the end, since the while loop will never terminate anyway? \$\endgroup\$
    – hyper-neutrino
    Jun 27 at 22:32
  • 1
    \$\begingroup\$ No, while 1: is not allowed, unfortunately. However, the space can be removed. Thanks for the tip. \$\endgroup\$
    – Pyautogui
    Jun 28 at 2:42
  • 1
    \$\begingroup\$ do you actually need to assign to a? can you just use readChar(stdin) in the condition? \$\endgroup\$
    – Jo King
    Jun 28 at 3:37
  • \$\begingroup\$ Oh, right. Sorry, stupid of me. \$\endgroup\$
    – Pyautogui
    Jun 28 at 20:07
0
\$\begingroup\$

n/t/roff, 27 bytes.

.de a
\\$1
.if \\$1 .a 1
..

Defines a macro a.

Usage

.a 0
.a 1

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ No wait I need to modify to accept from STDIN, not from macro argument. \$\endgroup\$ Aug 19 at 2:52
  • \$\begingroup\$ How the hell can I do so? \$\endgroup\$ Aug 19 at 12:13
0
\$\begingroup\$

Pure Bash, 32 + 1 = 33 bytes

The 1-byte penalty is for filename specification; the following program must be saved as x.

((x))||read x
echo $x
((x))&&. x

Try it online!

Tracing how it works

  • The variable x is not initialized.
  • ((x)) evaluates to ((0)): command is failure.
  • So read x is done.
  • Then echo $x.
  • Final line is for branching.
    • If x is zero, the program ends.
    • If x is one, . x is done.
      • It loads the file x: the program itself.
      • First command evaluates to ((1)) so no more reading.
      • Then rest two lines are executed and it loops forever.
\$\endgroup\$
0
\$\begingroup\$

INTERCAL, 41 bytes

Based on answer by @unrelatedstring. If you don't care for errors, won't you guys golf off the program easily?

DOWRITEIN.1DOCOMEFROM.1(1)PLEASEREADOUT.1

Try it online!

\$\endgroup\$
-1
\$\begingroup\$

Ruby, 22 bytes

->(n){n==0?0:loop{p1}}
\$\endgroup\$
-2
\$\begingroup\$

Python 2, 43 42 41 bytes

x=raw_input();print x
while x>'0':print x

Got a syntax error at the while loop if I tried to run it as a one-liner. Not really sure why.

EDIT: Knocked off a semicolon.

Knocked off ANOTHER semicolon that I put on there because I've been looking at C too much lately.

\$\endgroup\$
1
  • 3
    \$\begingroup\$ You could use x=input() and then x>0 \$\endgroup\$
    – Cyoce
    Sep 20 '16 at 17:41
1
11 12 13 14
15

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