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 <list>
#include <stack>
#include <limits.h>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <memory>
#include <regex>
#include <functional>
using namespace std;

int input[7][7];
int main() {
    int sum = 0;
    for (size_t i = 0; i < 3; i++)
    {
        for (size_t j = 0; j < 3; j++)
        {
            cin >> input[i][j];
            sum += input[i][j];
        }
    }
    if (sum % 3 != 0)
    {
        cout << "No" << endl;
        return 0;
    }
    int local_sum = input[0][0] + input[1][1] + input[2][2];
    if ((local_sum) % 3 != 0 || local_sum != sum / 3)
    {
        cout << "No" << endl;
        return 0;
    }
    local_sum = input[0][1] + input[1][2] + input[2][0];
    if (local_sum % 3 != 0 || local_sum != sum / 3)
    {
        cout << "No" << endl; local_sum;
        return 0;
    }

    local_sum = input[0][2] + input[1][1] + input[2][2];
    if (local_sum % 3 != 0 || local_sum != sum / 3)
    {
        cout << "No" << endl; local_sum;
        return 0;
    }

    local_sum = input[1][0] + input[2][1] + input[1][2];
    if (local_sum % 3 != 0 || local_sum != sum / 3)
    {
        cout << "No" << endl; local_sum;
        return 0;
    }

    local_sum = input[2][0] + input[1][1] + input[0][2];
    if (local_sum % 3 != 0 || local_sum != sum / 3)
    {
        cout << "No" << endl; local_sum;
        return 0;
    }

    cout << "Yes" << endl;
    return 0;
}