You are given a sequence of n integers a_1,   a_2,   ...,   a_n.

Does there exist a sequence of n integers b_1,   b_2,   ...,   b_n such that the following property holds?

  * For each 1 ≤ i ≤ n, there exist two (not necessarily distinct) indices j and k (1 ≤ j,   k ≤ n) such that a_i = b_j - b_k. 

Input

The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 10).

The second line of each test case contains the n integers a_1,   ...,   a_n (-10^5 ≤ a_i ≤ 10^5).

Output

For each test case, output a line containing YES if a sequence b_1,   ...,   b_n satisfying the required property exists, and NO otherwise.

Example

Input


5
5
4 -7 -1 5 10
1
0
3
1 10 100
4
-3 2 10 2
9
25 -171 250 174 152 242 100 -205 -258


Output


YES
YES
NO
YES
YES

Note

In the first test case, the sequence b = [-9,   2,   1,   3,   -2] satisfies the property. Indeed, the following holds: 

  * a_1 = 4 = 2 - (-2) = b_2 - b_5; 
  * a_2 = -7 = -9 - (-2) = b_1 - b_5; 
  * a_3 = -1 = 1 - 2 = b_3 - b_2; 
  * a_4 = 5 = 3 - (-2) = b_4 - b_5; 
  * a_5 = 10 = 1 - (-9) = b_3 - b_1. 



In the second test case, it is sufficient to choose b = [0], since a_1 = 0 = 0 - 0 = b_1 - b_1.

In the third test case, it is possible to show that no sequence b of length 3 satisfies the property.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;

void debug_out(){
    cerr << endl;
}
template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T){
    cerr << ' ' << H;
    debug_out(T...);
}
#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 55
#endif

const int N = 15;
int a[N], b[N], n, flag;
void dfs(int p){
    if(flag) return;
    if(p == n + 1){
        gp_hash_table<int, int> g;
        for(int i = 1; i <= n; i++) g[b[i]] = 1;
        for(int i = 1; i <= n; i++){
            if(g.find(a[n] - b[i]) != g.end() || g.find(a[n] + b[i]) != g.end()){
                flag = 1;
                break;
            }
        }
        return;
    }
    for(int i = 1; i < p; i++){
        b[p] = a[p - 1] + b[i];
        dfs(p + 1);
        b[p] = b[i] - a[p - 1];
        dfs(p + 1);
    }
}
int main() {
#ifdef local
    freopen("../in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while(T--){
        flag = 0;
        cin >> n;
        for(int i = 1; i <= n; i++) cin >> a[i];
        b[1] = 0;
        dfs(2);
        if(flag) cout << "YES\n";
        else cout << "NO\n";
    }
    return 0;
}