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;
set<int> s;
int arr[15], x[200005];
int get_sum(int mask, int n) {
  int ret = 0;
  for (int i = 0; i < n; i++) {
    if (mask & (1 << i)) ret += arr[i];
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    s.clear();
    for (int j = 0; j < 2e5 + 5; j++) {
      x[j] = 0;
    }
    for (int j = 0; j < n; j++) {
      cin >> arr[j];
      x[int(arr[j] + 1e5)] = j + 1;
      s.insert(arr[j]);
    }
    if (s.size() < n) {
      cout << "YES\n";
      continue;
    }
    int b1 = 0;
    for (int mask1 = 0; mask1 < (1 << n); mask1++) {
      for (int mask2 = 0; mask2 < (1 << n); mask2++) {
        if (mask1 & mask2) continue;
        int a = get_sum(mask1, n) - get_sum(mask2, n);
        if (a > 1e5) continue;
        if (!x[int(a + 1e5)]) continue;
        if (!(mask1 & (1 << (x[int(a + 1e5)] - 1))) &&
            !(mask2 & (1 << (x[int(a + 1e5)] - 1))))
          b1 = 1;
      }
    }
    if (b1)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
