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>
#define PB push_back;
#define MP make_pair;

typedef long long ll;

using namespace std;

void inpt(int n, vector <int> &v){
  for(int i=0;i<n;i++){
    int temp=0;
    cin>>temp;
    v.push_back(temp);
  }
}

bool is_prime(int n){
  int i=0,c=0;
  for(i=2;i<=(n/2);i++){
    if((n%i)==0){
      c=1;
      break;
    }
  }
  if(c==0){
    return true;
  }
  else{
    return false;
  }
}

void solve(){
  int n=0,i=0;
  cin>>n;
  if(n<2020){
    cout<<"NO"<<endl;
  }
  else if(n>=2020){
    if(n%2020==0){
      cout<<"YES"<<endl;
    }
    else if(n%2021==0){
      cout<<"yes"<<endl;
    }
    else if((n%2020!=0)&&(n%2021!=0)){
      i=0;
      while(n>=4041){
        n-=2021;
        if(n%2020==0){
          i=1;
          break;
        }
      }
      if(i==1){
        cout<<"yes"<<endl;
      }
      else if(i==0){
        cout<<"no"<<endl;
      }
    }
  }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("inputf.txt", "r", stdin);
    freopen("outputf.txt", "w", stdout);
#endif
    int t=0;
    cin>>t;
    while(t--){
      solve();
    }
}