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;
const double EPS = 1e-9;
const long long INF = 2e9, MOD = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long ind = 0; ind < n; ind++) {
      vector<long long> tmp = a;
      tmp.erase(tmp.begin() + ind);
      if (a[ind] == 0) {
        goto l;
      }
      for (long long i = 0; i < (1 << (n - 1)); i++) {
        long long s = 0;
        vector<long long> left;
        for (long long j = 0; j < n - 1; j++) {
          if (i & (1 << j)) {
            s += tmp[j];
          } else {
            left.push_back(tmp[j]);
          }
        }
        long long cur;
        long long num = (long long)left.size();
        for (long long j = 0; j < (1 << num); j++) {
          cur = 0;
          for (long long x = 0; x < num; x++) {
            if (j & (1 << x)) {
              cur += left[x];
              if (abs(cur - s) == abs(a[ind]) || abs(cur + s) == abs(a[ind])) {
                goto l;
              }
            }
          }
        }
      }
    }
    cout << "NO"
         << "\n";
    continue;
  l:
    cout << "YES"
         << "\n";
  }
  return 0;
}
