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 mod 998244353
#define modc 1000000007
     
const long double PI = 3.141592653589793236L;
     
typedef long long int ll;
typedef long double ld;
     
using namespace std;

ll fast_power(ll a, ll b, ll m)//a^b
{
    if(b==0) return 1;
    if(b==1) return a;

    ll temp=fast_power(a,b/2,m);

    if(b%2==0) return (temp*temp)%m;
    return ((temp*temp)%m*a)%m;
}

ll gcd(ll a, ll b)
{
    if (b == 0) return a;
    return gcd(b, a % b); 
}

ll lcm(ll a,ll b)
{
    return (a*b)/gcd(a,b);
}

ll sum_of_digits(ll n)
{
    ll sum=0;
    while(n>0)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}

void solve()
{
    //check t
    ll n;
    cin>>n;
    if(n<2020) cout<<"NO\n";
    else if(n%2020==0 or n%2021==0) cout<<"YES\n";
    else
    {
        while(n>2020)
        {
            n-=2020;
            if(n%2021==0) 
            {
                cout<<"YES\n";
                return;
            }
        }
        cout<<"NO\n";
    }
}
     
int main()
{   
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    ll t;
    t=1;
    cin>>t;
    while(t--)
    {   
        solve();
    }
    return 0;
}