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 <bits/stdc++.h>
using namespace std;
using i64 = long long;
using P = pair<int, int>;
using vec = vector<int>;
using mat = vector<vector<int>>;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define endl "\n"

constexpr int MOD = 1000000007;
constexpr int INF = 1001001001;

constexpr bool is_multiple = true;

bool table[1000010];

void make_table() {
    table[0] = true;

    for (int i = 2020; i <= 1000000; i++) {
        if (table[i-2020]) table[i] = true;
        if (i >= 2021 && table[i-2021]) table[i] = true;
    }
}

void solve() {
    int n;
    cin >> n;

    if (table[n]) cout << "YES" << endl;
    else cout << "NO" << endl;
}

int main() {
    int t = 1;
    if (is_multiple) cin >> t;
    make_table();
    while (t--) solve();
    return 0;
}
