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
//#define LOCAL
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const static int inf = 0x3f3f3f3f;
const static ll INF = 0x3f3f3f3f3f3f3f3f;
const static int maxn = 1e5+10;

int main() {
	#ifdef LOCAL
	freopen("1.in", "r", stdin);
	freopen("1.out", "w", stdout);
	#endif
	int T;
	cin >> T;
	while(T--) {
		ll n;
		cin >> n;
		int a = n / 2020;
		int b = n % 2020;
		if(b <= a && a >= 1) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}
