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 <iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<utility>
#include<cstring>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define ii pair<int,int>
#define vi vector<int>
#define MS 1000001
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define lli long long int
#define ulli unsigned long long int
static lli INF=1e9+7;
using namespace std;
 int dp[MS];
 bool solve(int n){
 	if(n==0)
 		return true;
 	if(n<2021 && n<2020)
 		return false;
 	if(dp[n]!=-1)
 		return dp[n];
 	return dp[n]=solve(n-2020)||solve(n-2021);
 }
int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
   int t;
   cin>>t;
   memset(dp,-1,sizeof dp);
   while(t--){
   	int n;
   	cin>>n;
   bool ans=solve(n);
   if(ans)
   	cout<<"YES";
   else cout<<"NO";
   cout<<endl;
}
   
}