#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MOD = 1000000007;
const long double PI = 3.141592653589793238;
const long long pi = 31415926;
const long long inf = 1000000000000000000;
const long long small_inf = INT_MAX;
const int N = 300002;
int power(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
int add(int x, int y) { return (x + y) % MOD; }
int mult(int x, int y) { return (x * y) % MOD; }
void init() {}
void solve() {
  int n;
  cin >> n;
  std::vector<int> v(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < power(2, n); i++) {
    std::vector<int> v1;
    if (i == 0) continue;
    for (int j = 0; j < n; j++) {
      if ((i & (1 << j))) v1.push_back(v[j]);
    }
    {
      int n = v1.size();
      for (int i = 0; i < power(2, n); i++) {
        int x = 0;
        for (int j = 0; j < n; j++) {
          if (i & (1 << j)) {
            x += v1[j];
          } else
            x -= v1[j];
        }
        if (x == 0) {
          cout << "YES" << endl;
          return;
        }
      }
    }
  }
  cout << "NO\n";
}
void clear_global() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t23 = 1, tt23 = 1;
  cin >> t23;
  init();
  while (tt23 <= t23) {
    solve();
    tt23++;
    clear_global();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MOD = 1000000007;
const long double PI = 3.141592653589793238;
const long long pi = 31415926;
const long long inf = 1000000000000000000;
const long long small_inf = INT_MAX;
const int N = 300002;
int power(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
int add(int x, int y) { return (x + y) % MOD; }
int mult(int x, int y) { return (x * y) % MOD; }
void init() {}
void solve() {
  int n;
  cin >> n;
  std::vector<int> v(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < power(2, n); i++) {
    std::vector<int> v1;
    if (i == 0) continue;
    for (int j = 0; j < n; j++) {
      if ((i & (1 << j))) v1.push_back(v[j]);
    }
    {
      int n = v1.size();
      for (int i = 0; i < power(2, n); i++) {
        int x = 0;
        for (int j = 0; j < n; j++) {
          if (i & (1 << j)) {
            x += v1[j];
          } else
            x -= v1[j];
        }
        if (x == 0) {
          cout << "YES" << endl;
          return;
        }
      }
    }
  }
  cout << "NO\n";
}
void clear_global() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t23 = 1, tt23 = 1;
  cin >> t23;
  init();
  while (tt23 <= t23) {
    solve();
    tt23++;
    clear_global();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```