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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
#define int int64
#define rep(i, n) for(int i=0; i<n; i++)
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define SORT(x) sort(x.begin(), x.end())
#define GSORT(x) sort(x.begin(), x.end(), greater<int>())
#define mk make_pair
#define fi first
#define se second
#define pb push_back
#define ALL(x) x.begin(), x.end()
#define V(T) vector<T>
typedef pair<int, int> P;
typedef pair<P, P> PP;
typedef vector<int> vi;
typedef vector<vi> vvi;
int max(int a, int b) {if(b>a) return b; else return a;}
int min(int a, int b) {if(b<a) return b; else return a;}


signed main() {
    int N, A;
    cin >> N >> A;
    
    if(N%500 <= A) cout << "Yes";
    else cout << "No";
    cout << endl;

    return 0;
}