Problem:
Score: 100 points

Problem Statement

E869120 has A 1-yen coins and infinitely many 500-yen coins.

Determine if he can pay exactly N yen using only these coins.

Constraints

N is an integer between 1 and 10000 (inclusive).

A is an integer between 0 and 1000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
A

Output

If E869120 can pay exactly N yen using only his 1-yen and 500-yen coins, print Yes; otherwise, print No.

Sample Input 1

2018
218

Sample Output 1

Yes

We can pay 2018 yen with four 500-yen coins and 18 1-yen coins, so the answer is Yes.

Sample Input 2

2763
0

Sample Output 2

No

When we have no 1-yen coins, we can only pay a multiple of 500 yen using only 500-yen coins. Since 2763 is not a multiple of 500, we cannot pay this amount.

Sample Input 3

37
514

Sample Output 3

Yes

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>

int main() 
{
	int N;
	int a[100];
	int A = 0;
	int B = 0;
	int i, j;
	
	scanf("%d",&N);
	for(i=0; i<N; i++){
		scanf("%d",&a[i]);
	}
	
	for(i=0; i<N; i++){
		for(j=N-1; j>i; j--){
			int temp = a[j];
			if( a[j-1] < a[j] ) {
			a[j] = a[j-1];
			a[j-1] = temp;
			}
		}
	}
	for(i=0; i<N; i++){
		if(i%2) B += a[i];
		else A += a[i];
	}
	printf("%d",A-B);
	
	return 0;
}