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
/*
@author: Saurav Sihag
*/

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ull unsigned long long
#define ff first
#define ss second
#define lld long double

template <class T>
void print(vector<T>& v){
    for(T x:v) cout<<x<<" ";
    cout<<"\n";
}


void inp(vector<ll>& v, int n){
    for (int i=0; i<n; ++i) scanf("%lld", &v[i]);
}

// const int nax=1e+5+5;
// const int inf=1e+9+5;
// const ll mod=1e+9+7;
// const ll mod=998244353;
// const ll nax=1e+16+5;
// const int dx[4]={0,0,-1,1};
// const int dy[4]={1,-1,0,0};



void sol(){
    ll n;
    cin>>n;
    ll x=n/2020, y=n%2020;
    if (x>=y) printf("YES\n");
    else printf("NO\n");
    
    return;
}



int main() {
    int t;
    // t=1;
    cin>>t;
    t++;
    // std::cout << std::setprecision(20);
    for (int i=1; i<t; ++i){ 
        // printf("Case #%d: ", i);
        sol();
    }
    return 0;
}