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
#include <iostream>	
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <chrono>
#include <tuple>
#include <random>
#include <cmath>
#include <iomanip>
#include <random>
#include <unordered_map>
#define int long long
#define Nado cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0); //setlocale(LC_ALL, "Russian");
using namespace std;
mt19937 rnd(42);
signed main() {
	Nado;
	int ttt;
	cin >> ttt;
	for (int uuu = 1; uuu <= ttt; ++uuu) {
		int n; 
		cin >> n;
		if (n < 2020) {
			cout << "NO";
		}
		else {
			int k = n / 2020;
			if (n - k * 2020 >= 0 && n - k * 2020 <= k) {
				cout << "YES";
			}
			else {
				cout << "NO";
			}
		}
		cout << endl;
	}
	return 0;
}