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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
 
typedef pair<ll,ll> pairs;
vector<pairs> p;
bool pairCompare(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.first < secondElof.first;
}
#define MAX_N 1000100
bool x[MAX_N];
ll num[MAX_N];
ll fibl[MAX_N]={0};
 
ll fib(ll a){
    if (fibl[a]!=0)return fibl[a];
    if (a==0){
        return 0;
    }else if (a==1){
        return 1;
    }
    return fibl[a]=fib(a-1)+fib(a-2);
}
 
ll eratosthenes(ll n) {
    int p = 0;
    for (ll i=0; i<=n; ++i) x[i] = true;
    x[0] = x[1] = false;
    for(int i=2; i<=n; ++i) {
        if(x[i]) {
            p++;
            for(int j=2*i; j<=n; j+=i) x[j] = false;
        }
        
        num[i] = p;
    }
    return p;
}
 
ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}
 
ll keta(ll N){
    int tmp{};
    while( N > 0 ){
        tmp += ( N % 10 );
        N /= 10;
    }
    N = tmp;
    return N;
}

int main(){
    ll x,a,b;
    cin >>x>>a>>b;
    
    cout << (x-a)%b <<endl;
    
    return 0;
}