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 pb          push_back
#define mp          make_pair
#define ll          long long int
//#define int         long long int
#define pll         pair<ll,ll> 
#define hmap        unordered_map
#define M           1000000007
#define MOD         998244353
#define fast        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define test        ll tt; cin>>tt; while(tt--)
#define inp(a,n)    int a[n]; for(int i=0;i<n;i++)  cin>>a[i];
#define inpl(a,n)   ll a[n];  for(ll i=0;i<n;i++)   cin>>a[i];
#define out(a,n)    for(int i=0;i<n;i++) cout<<a[i]<<" "; cout<<endl;
#define outl(a,n)   for(ll i=0;i<n;i++)  cout<<a[i]<<" "; cout<<endl;
#define outv(v)     for(int i=0;i<v.size();i++) cout<<v[i]<<" ";  cout<<endl;
#define all(s)      (s).begin(),(s).end()

// ll n = 1000001;
// bool a[1000001]={};
// vector<ll> v;
// void prime()
// {
//       for(ll i=0;i<n;i++)
//             a[i]=0;
            
//       for(ll i=3;i<n;i+=2)
//             a[i]=1;
//       a[2]=1;
//       v.pb(2);
      
//       for(ll i=3;i<n;i+=2)
//       {
//             if(a[i]==1)
//             {
//                   v.pb(i);
//                   for(ll j=i*i ; j<n ; j += i)
//                         a[j]=0;
//             }
//       }
// }


int main()
{
      fast
      test
      {
            ll n;
            cin>> n;
            
            ll c = n/2020;
            ll d = n - c*2020;
            
            if(n<2020 || (d>c))
                  cout<<"NO";
            else
                  cout<<"YES";
            cout<<endl;
            
      }
}
