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>

void sort(int a[], int n)
{
	int i,j,temp;

	for(i=0;i<n-1;i++){
		for(j=i;j<n-1;j++){
			if(a[j+1]>a[j]){
				temp = a[j];
				a[j] = a[j+1];
				a[j+1] = temp;
			}
		}
	}
}

int main(void)
{
	unsigned int N,a[N],i=0;
	unsigned int Alice=0,Bob=0;

	scanf("%d",&N);

	for(i=0;i<N;i++){
		scanf("%d ",&a[i]);
		i++;
	}

	sort(a , N);
	
	for(i=0;i<=N/2;i++){
		if(2*i<=N){
			Alice += a[2*i];
		}
		if(2*i+1<=N){
			Bob += a[2*i+1];
		}
	}

	printf("%d\n",Alice-Bob);

	return 0; 
}