#include <bits/stdc++.h>
using namespace std;
void sol(long long int x, long long int y, long long int ind,
         long long int &flag, vector<long long int> &a) {
  if (x == y && x != 0) {
    flag = 1;
    return;
  }
  if (ind == a.size()) return;
  sol(x + a[ind], y, ind + 1, flag, a);
  if (flag == 1) return;
  sol(x, y, ind + 1, flag, a);
  if (flag == 1) return;
  sol(x, y + a[ind], ind + 1, flag, a);
  if (flag == 1) return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t, l, flag, od, ev, sum, ma, mi, i, j, k, n, m, ans, tot, count,
      x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long int> a;
    for (i = 0; i < n; i++) {
      cin >> x;
      a.push_back(abs(x));
    }
    sort(a.begin(), a.end());
    if (a[0] == 0)
      cout << "YES\n";
    else if (n == 1)
      cout << "NO\n";
    else if (n == 2) {
      if (a[0] == a[1])
        cout << "YES\n";
      else
        cout << "NO\n";
    } else {
      flag = 0;
      for (i = 0; i < n - 1; i++) {
        if (a[i] == a[i + 1]) {
          flag = 1;
          break;
        }
      }
      if (flag == 1)
        cout << "YES\n";
      else {
        for (i = 0; i < n; i++) {
          for (j = 0; j < n; j++) {
            for (k = j + 1; k < n; k++) {
              if (i != j && i != k) {
                if (a[j] + a[k] == a[i] || a[k] - a[j] == a[i]) {
                  flag = 1;
                  break;
                }
              }
            }
            if (flag == 1) break;
          }
          if (flag == 1) break;
        }
        if (flag == 1)
          cout << "YES\n";
        else {
          sol(0, 0, 0, flag, a);
          if (flag == 1)
            cout << "YES\n";
          else
            cout << "NO\n";
        }
      }
    }
  }
  return 0;
}

### 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>
using namespace std;
void sol(long long int x, long long int y, long long int ind,
         long long int &flag, vector<long long int> &a) {
  if (x == y && x != 0) {
    flag = 1;
    return;
  }
  if (ind == a.size()) return;
  sol(x + a[ind], y, ind + 1, flag, a);
  if (flag == 1) return;
  sol(x, y, ind + 1, flag, a);
  if (flag == 1) return;
  sol(x, y + a[ind], ind + 1, flag, a);
  if (flag == 1) return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t, l, flag, od, ev, sum, ma, mi, i, j, k, n, m, ans, tot, count,
      x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long int> a;
    for (i = 0; i < n; i++) {
      cin >> x;
      a.push_back(abs(x));
    }
    sort(a.begin(), a.end());
    if (a[0] == 0)
      cout << "YES\n";
    else if (n == 1)
      cout << "NO\n";
    else if (n == 2) {
      if (a[0] == a[1])
        cout << "YES\n";
      else
        cout << "NO\n";
    } else {
      flag = 0;
      for (i = 0; i < n - 1; i++) {
        if (a[i] == a[i + 1]) {
          flag = 1;
          break;
        }
      }
      if (flag == 1)
        cout << "YES\n";
      else {
        for (i = 0; i < n; i++) {
          for (j = 0; j < n; j++) {
            for (k = j + 1; k < n; k++) {
              if (i != j && i != k) {
                if (a[j] + a[k] == a[i] || a[k] - a[j] == a[i]) {
                  flag = 1;
                  break;
                }
              }
            }
            if (flag == 1) break;
          }
          if (flag == 1) break;
        }
        if (flag == 1)
          cout << "YES\n";
        else {
          sol(0, 0, 0, flag, a);
          if (flag == 1)
            cout << "YES\n";
          else
            cout << "NO\n";
        }
      }
    }
  }
  return 0;
}

```