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<cmath>    
#include<queue>    
#include<vector>    
#include<cstdio>    
#include<ctime>   
#include<cstring>    
#include<iostream>    
#include<algorithm>    
#define INF 0x3f3f3f3f
#define MAXN 100000
using namespace std;
int read()
{
    int f=1,x=0;char s=getchar();   
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}  
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    return x*f;
}
int main() 
{
	int x=read(),a=read(),b=read();
	printf("%d\n",(x-a)%b);
    return 0; 
}