Problem:
Score : 100 points

Problem Statement

You went shopping to buy cakes and donuts with X yen (the currency of Japan).

First, you bought one cake for A yen at a cake shop.
Then, you bought as many donuts as possible for B yen each, at a donut shop.

How much do you have left after shopping?

Constraints

1 \leq A, B \leq 1 000

A + B \leq X \leq 10 000

X, A and B are integers.

Input

Input is given from Standard Input in the following format:

X
A
B

Output

Print the amount you have left after shopping.

Sample Input 1

1234
150
100

Sample Output 1

84

You have 1234 - 150 = 1084 yen left after buying a cake.
With this amount, you can buy 10 donuts, after which you have 84 yen left.

Sample Input 2

1000
108
108

Sample Output 2

28

Sample Input 3

579
123
456

Sample Output 3

0

Sample Input 4

7477
549
593

Sample Output 4

405

Solution in JavaScript (node.js v5.12) (judge verdict: Wrong Answer)



var Release = function () {
    var fs = require('fs');
    var stdin = function (text) {
        if (text) console.log(text);
        return fs.readFileSync('/dev/stdin', 'utf8');
    }
    return stdin;
};


var debug = false;

var stdin = debug ? Debug() : Release();
var input = stdin().split('\n');
var NM = input.shift().split(' ').map(x => parseInt(x));
var N = NM[0];
var M = NM[1];
var info = input.map(x => x.split(' ').map(y => parseInt(y)));

var h = true;
for (var i = 0; i < M; i++) {
    var t = info.pop();
    var L = t[0];
    var R = t[1];
    var D = t[2];

    var l = info.filter(x => x[0] === L);
    for(var j=0; j<l.length; j++) {
        var ll = info.filter(x => x[0] === l[j][1] && x[1] === R).concat(info.filter(
            x => x[1] === l[j][1] && x[0] === R
        )).filter(x => (l[j][2] + x[2] - D) !== 0);
        if (ll.length > 0) {
            h = false;
            break;
        }
    }
    if (!h) break;

    
    var l = info.filter(x => x[1] === L);
    for(var j=0; j<l.length; j++) {
        var ll = info.filter(x => x[0] === l[j][0] && x[1] === R).concat(info.filter(
            x => x[1] === l[j][0] && x[0] === R
        )).filter(x => (l[j][2] + x[2] - D) !== 0);
        if (ll.length > 0) {
            h = false;
            break;
        }
    }
    if (!h) break;

}


console.log(h ? 'Yes' : 'No');