#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7;
int t, dp[15][N * 20], n, a[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> t;
  int MX = 1e6;
  while (t--) {
    cin >> n;
    int first = 0;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < (1 << n); i++) {
      int sum = 0;
      vector<int> x, y;
      for (int j = 0; j < n; j++) {
        if ((1 << j) & i) {
          x.push_back(j);
        } else
          y.push_back(j);
      }
      int cnt = x.size();
      for (int j = 0; j < (1 << cnt); j++) {
        int sum = 0;
        for (int k = 0; k < cnt; k++) {
          if ((1 << k) & j) {
            sum += a[x[k]];
          } else
            sum -= a[x[k]];
        }
        for (int k = 0; k < y.size(); k++) {
          dp[y[k]][sum + MX]++;
        }
        if (sum == 0 && i) first = 1;
      }
    }
    a[n] = 0;
    for (int i = 0; i < n; i++) {
      if (dp[i][a[i] + MX] > 1 || dp[i][-a[i] + MX] > 1) first = 1;
    }
    for (int i = 0; i <= 2 * MX; i++)
      for (int j = 0; j <= n; j++) dp[j][i] = 0;
    if (first)
      cout << "YES"
           << " ";
    else
      cout << "NO"
           << " ";
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7;
int t, dp[15][N * 20], n, a[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> t;
  int MX = 1e6;
  while (t--) {
    cin >> n;
    int first = 0;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < (1 << n); i++) {
      int sum = 0;
      vector<int> x, y;
      for (int j = 0; j < n; j++) {
        if ((1 << j) & i) {
          x.push_back(j);
        } else
          y.push_back(j);
      }
      int cnt = x.size();
      for (int j = 0; j < (1 << cnt); j++) {
        int sum = 0;
        for (int k = 0; k < cnt; k++) {
          if ((1 << k) & j) {
            sum += a[x[k]];
          } else
            sum -= a[x[k]];
        }
        for (int k = 0; k < y.size(); k++) {
          dp[y[k]][sum + MX]++;
        }
        if (sum == 0 && i) first = 1;
      }
    }
    a[n] = 0;
    for (int i = 0; i < n; i++) {
      if (dp[i][a[i] + MX] > 1 || dp[i][-a[i] + MX] > 1) first = 1;
    }
    for (int i = 0; i <= 2 * MX; i++)
      for (int j = 0; j <= n; j++) dp[j][i] = 0;
    if (first)
      cout << "YES"
           << " ";
    else
      cout << "NO"
           << " ";
  }
}

```