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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int main() {
  char *a = malloc(sizeof(char) * 4);
  char *b = malloc(sizeof(char) * 4);
  char *c = malloc(sizeof(char) * 4);
  char *x = malloc(sizeof(char) * 4);
  int na, nb, nc, nx;
  
  fgets(a, 3, stdin);
  fgets(b, 3, stdin);
  fgets(c, 3, stdin);
  fgets(x, 6, stdin);

  na = atoi(a);
  nb = atoi(b);
  nc = atoi(c);
  nx = atoi(x);

  printf("%d\n", na);
  printf("%d\n", nb);
  printf("%d\n", nc);
  printf("%d\n", nx);
  
  int n = 0;
  
  for(int i = 0; i <= na; i++) {
    for(int j = 0; j <= nb; j++ ) {
      for(int k = 0; k <= nc; k++) {
	if((500*i) + (100*j) + (50*k) == nx)
	  n++;
      }
    }
  }

  printf("%d\n", n);
  
  return 0;
}