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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<iterator>
#include<vector>
#include<string>
#include<map>
using namespace std;


#define mod(cans) cans%10**9+7
#define cin_pl(x,y) for(int i=0;i<x;i++){cin>>y[i];}
#define pl_twin(x,y,z) for(int i=0;i<x;i++){cin>>y[i]>>z[i];}
#define debug(x) cout << "Debug" << x << "\n"
#define rep(i,n) for(int i=0;i<n;i++)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define st(s) size_t size = s.length()
#define prians(Answer) cout<<Answer<<"\n"
#define make_pair m_p;
#define com_div(x,Co) for(int i=1;i<x+1;i++){if(x%i==0){Co++;}}
#define elcon else{continue;}

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vector<int>> matrix;
typedef pair<int,int> pr;
typedef greater<int> gi;

int main(void)
{
    int n,a;
    cin>>n>>a;
    int tmp=0;
    for(int i=500;i<10000;i+=500)
    {
        if(n%i<a)
        {
            tmp=1;
            break;
        }
    }
    tmp==1?cout<<"Yes":cout<<"No";
}