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>
using namespace std;
signed main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n + 1);
    set<long long> s;
    for (long long i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    bool zero = false;
    for (long long i = 1; i <= n; i++) {
      s.insert(arr[i]);
      if (arr[i] == 0) zero = true;
    }
    if (s.size() < n || zero) {
      cout << "YES\n";
      continue;
    }
    bool f = false;
    for (long long k = 0; k < n; k++) {
      vector<long long> arr2;
      for (long long i = 0; i < n; i++) {
        if (i == k) continue;
        arr2.push_back(arr[i + 1]);
      }
      vector<long long> q;
      vector<long long> q2;
      q.push_back(0);
      for (long long i = 0; i < n - 1; i++) {
        for (long long nxt : q) {
          q2.push_back(nxt + arr2[i]);
          q2.push_back(nxt);
          q2.push_back(nxt - arr2[i]);
        }
        swap(q, q2);
        q2.clear();
      }
      for (long long nxt : q) {
        if (nxt == arr[k + 1]) f = true;
      }
    }
    if (f)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
