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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

#define rep(i,a,b) for(ll i=a;i<b;i++)
#define rrep(i,b,a) for(int i=b;i>=a;i--)
#define fori(a) for(auto i : a )
#define all(a) begin(a), end(a)
#define set(a,b) memset(a,b,sizeof(a))
#define sz(a) a.size()
double pi=acos(-1);
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PF push_front //deque
// #define mp make_pair
#define pq priority_queue
const ll mod=1000000007;
#define f first
#define s second
#define pii pair< ll, ll >
#define vi vector<int>
#define vpii vector<pii>
#define debug(v) for(auto i:v) cout<<i<<" ";
#define tc int t; cin >> t; while(t--)

using namespace boost::multiprecision;
using namespace std;
void optimizeIO(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}


const int N=1000005;
ll fact[N],invfact[N];
ll power(ll x,ll y){
  if(y<=0) return 1LL;
  ll z=power(x,y/2);
  if(y%2) return (((z*z)%mod)*x)%mod;
  return (z*z)%mod;
}

void pre(){
  fact[0]=invfact[0]=invfact[1]=fact[1]=1;
  rep(i,2,N) fact[i]=(i*fact[i-1])%mod;
  rep(i,2,N) invfact[i]=(invfact[i-1]*power(i,mod-2))%mod;
}
ll nCr(ll n,ll k){ return (((fact[n]*invfact[n-k])%mod)*invfact[k])%mod; }



const int N1=1e6+1;
vector<int> isprime(N1,1),prime;
void seive(){
  rep(i,2,sqrt(N1)+1){
      if(isprime[i]){
        for(int j=i*i;j<N1;j+=i) isprime[j]=0;
        prime.pb(i);
      }
  }
  rep(i,sqrt(N1)+1,N1) if(isprime[i]) prime.pb(i);
}



struct dsu {
    vector<int> par, rank;
    dsu(int n): par(n+1), rank(n+1) {
        for (int i = 0; i <= n; i++) {
            par[i] = i;
            rank[i]= 1;
        }
    }
    int root(int a) {
        if (a == par[a]) return a;
        return par[a] = root(par[a]);
    }
    void merge(int a, int b) {
        a = root(a);
        b = root(b);
        if (a == b) return;
        if (rank[a] > rank[b]) swap(a, b);
        par[b] = a;
    }
    set<int> parent(int n){
        set<int> s;
        for(int i=1;i<=n;i++){
            s.insert(root(i));
        }
        return s;
    }
};

void solve(){
  int n,a;
  cin>>n>>a;
  n=n%500;
  if(n<=a) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}

int main(){
    optimizeIO();
    int r=1;
    {solve();}
}