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 long long ll;
#define lp(i,n) for(ll i = 0;i < ll(n);i++)
#define rlp(i,n) for(ll i = ll(n);i >= 0;i--)
#define rg(i,a,b) for(ll i = ll(a);i <= ll(b);i++)
#define re return
#define pb push_back
#define sz(s) (ll)(s.size())
#define fx(n) fixed<<setprecision(n)
#define NeverGiveUp ios::sync_with_stdio(0), ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
#define F first
#define S second
/*freopen("input.txt","r", stdin);
	freopen("output.txt","w", stdout);*/
using namespace std;
/*ll mx = INT_MIN;
ll mn = INT_MAX;*/
/*ll xd,dx[4]={1,0,-1,0};
ll yd,dy[4]={0,1,0,-1};
bool valid(ll xd,llD:\Programing\MyPractise\Assuit session\main.cpp yd,ll n,ll m){return xd>-1&&xd<n&&yd>-1&&yd<m;}*/
const int sz = 2e6+5;
const ll md = 1e9+7;
//ll lcm(ll a,ll b){return a*b/__gcd(a,b);}

int main()
{
    NeverGiveUp
    ll t=1;
    cin>>t;
    while(t--){
        ll n,c=0;
        cin>>n;
        while(n>=2020){
            n-=2020;
            c++;
        }
        if(n<=c)cout<<"YES\n";
        else cout<<"NO\n";
    }
    re 0;
}
