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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
#include <string>
#include <limits.h>
#include <cmath>
#include <queue>
#include <map>
#include <iomanip>
#include <random>
#include <fstream>
#define rep(i, n) for (int i = 0; i < n; i++)
#define PI 3.14159265358979323846
#define MOD 1000000007
using namespace std;
using ll = long long;

const int MAX = 1000000;

ll fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++){
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k){
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll gcd(ll a, ll b)
{
  ll r;
  while ((r = a % b))
  {
    a = b;
    b = r;
  }
  return b;
}

template <class T> struct cord{
  T x,y;
};

template <class T> double dist(cord<T> a, cord<T> b){
  T dx=a.x-b.x;
  T dy=a.y-b.y;
  return sqrt(dx*dx+dy*dy);
}

int main()
{

  //cout << fixed << setprecision(10);
  
  int x,a,b; cin >> x >> a >> b;
  cout << x - a - (x-a)/b * b;

  return 0;
}