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> 
typedef unsigned long long ull;
typedef long long ll;
typedef unsigned int ui;
typedef long double ld;
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define SORT(v) sort(all(v))
 
ll binpow(ll a,ll b){
    if(b==0){
        return 1;
    }
    ll result=binpow(a,b/2);
    if(b%2){
        return a*result*result;
    }else{
        return result*result;
    }
}
 
using namespace std;

ll digit_sum(ll n){
    if(n<=9){
        return n;
    }else{
        return n%10+digit_sum(n/10);
    }
}


void sol(){
    int n;
    cin >> n;
    int count =n/2020;
    int remain=n%2020;
    if(remain<=count){
        cout << "YES" << endl;
    }else{
        cout  << "NO" << endl;
    }

}

int main()
{   
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll tttt;
    cin >> tttt;
    while(tttt--)
        sol();
 
    return 0;
}
