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;
#define int long long
bool isPowerOfTwo(int n)
{
   if(n==0)
   return false;
 
   return (ceil(log2(n)) == floor(log2(n)));
}
bool isPrime(int n) 
{ 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
  

    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
  
    for (int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
  
    return true; 
} 

int32_t main() {
    
    int t;cin>>t;
    
    
    while(t--)
    {
       int n;
       cin>>n;
       int rup=n/2020;
       int spen=n%2020;
       if(spen<=rup)
       {
           cout<<"YES"<<endl;
       }
       else
       {
           cout<<"NO"<<endl;
       }
        
        
    }
    

  
    

	return 0;
}