Polycarp remembered the 2020-th year, and he is happy with the arrival of the new 2021-th year. To remember such a wonderful moment, Polycarp wants to represent the number n as the sum of a certain number of 2020 and a certain number of 2021.

For example, if: 

  * n=4041, then the number n can be represented as the sum 2020 + 2021; 
  * n=4042, then the number n can be represented as the sum 2021 + 2021; 
  * n=8081, then the number n can be represented as the sum 2020 + 2020 + 2020 + 2021; 
  * n=8079, then the number n cannot be represented as the sum of the numbers 2020 and 2021. 



Help Polycarp to find out whether the number n can be represented as the sum of a certain number of numbers 2020 and a certain number of numbers 2021.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case contains one integer n (1 ≤ n ≤ 10^6) — the number that Polycarp wants to represent as the sum of the numbers 2020 and 2021.

Output

For each test case, output on a separate line: 

  * "YES" if the number n is representable as the sum of a certain number of 2020 and a certain number of 2021; 
  * "NO" otherwise. 



You can output "YES" and "NO" in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

Example

Input


5
1
4041
4042
8081
8079


Output


NO
YES
YES
YES
NO
// Codeforces :: Round 697 Div 3 :: B. New Year's Number
// jramaswami

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

bool solve(int x) {
    int sub = 2021;
    if (x > 0 && x % 2021 == 0) {
        return true;
    }

    while (x > 2021 && x % 2020 != 0) {
        x -= sub;
    }
    return x > 0 && x % 2020 == 0;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int testCases{0};
    cin >> testCases;
    for (int t{0}; t < testCases; ++t) {
        int N;
        cin >> N;
        if (solve(N)) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
    }
    return EXIT_SUCCESS;
}
