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 double ld;
#define int long long

#define gcd            __gcd
#define endl           "\n"
#define setbits(x)     __builtin_popcountll(x)
#define zrobits(x)     __builtin_ctzll(x)
#define mod            1000000007
#define mod2           998244353
#define maxe           *max_element
#define mine           *min_element
#define inf            1e18
#define pb             push_back
#define all(x)         x.begin(), x.end()
#define f              first
#define s              second
#define lb             lower_bound
#define ub             upper_bound
#define ins            insert
#define sz(x)          (int)(x).size()
#define mk             make_pair
#define deci(x, y)     fixed<<setprecision(y)<<x
#define w(t)           int t; cin>>t; while(t--)
#define nitin          ios_base::sync_with_stdio(false); cin.tie(nullptr)
#define PI             3.141592653589793238
#define mem0(x)        memset(x,0,sizeof x)
#define mem1(x)        memset(x,-1,sizeof x)
using namespace std;
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.f << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }

#ifdef NITIN
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif


void solve() {
    int n;
    cin>>n;
    if((n/2020)>=(n%2020)){
        cout<<"YES";
    }else{
        cout<<"NO";
    }
    cout<<endl;
}

int32_t main() {
    nitin;
w(t)
    solve();
}