#include <bits/stdc++.h>
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 10;
int T, n, a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = (0); i <= (n - 1); ++i) cin >> a[i];
    sort(a, a + n);
    bool ok = 0;
    for (int i = (0); i <= (n - 1); ++i) {
      if (a[i] == 0 || (i > 0 && a[i] == a[i - 1])) ok = 1;
    }
    for (int i = (0); i <= (n - 1); ++i) {
      for (int j = (0); j <= (n - 1); ++j)
        if (a[j] == -a[i]) {
          ok = 1;
          break;
        }
      for (int x = (0); x <= ((1 << n) - 1); ++x)
        if (!(x >> i & 1)) {
          for (int mask = x;; mask = (mask - 1) & x) {
            int sum = 0;
            for (int j = (0); j <= (n - 1); ++j)
              if (x >> j & 1) {
                if (mask >> j & 1)
                  sum += a[j];
                else
                  sum -= a[j];
              }
            if (sum == a[i]) {
              ok = 1;
              break;
            }
            if (mask == 0) break;
          }
        }
    }
    cout << (ok ? "YES\n" : "NO\n");
  }
  return 0;
}

### 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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 10;
int T, n, a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = (0); i <= (n - 1); ++i) cin >> a[i];
    sort(a, a + n);
    bool ok = 0;
    for (int i = (0); i <= (n - 1); ++i) {
      if (a[i] == 0 || (i > 0 && a[i] == a[i - 1])) ok = 1;
    }
    for (int i = (0); i <= (n - 1); ++i) {
      for (int j = (0); j <= (n - 1); ++j)
        if (a[j] == -a[i]) {
          ok = 1;
          break;
        }
      for (int x = (0); x <= ((1 << n) - 1); ++x)
        if (!(x >> i & 1)) {
          for (int mask = x;; mask = (mask - 1) & x) {
            int sum = 0;
            for (int j = (0); j <= (n - 1); ++j)
              if (x >> j & 1) {
                if (mask >> j & 1)
                  sum += a[j];
                else
                  sum -= a[j];
              }
            if (sum == a[i]) {
              ok = 1;
              break;
            }
            if (mask == 0) break;
          }
        }
    }
    cout << (ok ? "YES\n" : "NO\n");
  }
  return 0;
}

```