#include <bits/stdc++.h>
using namespace std;
set<int> s;
set<int> s2;
int main() {
  int ttt;
  cin >> ttt;
  while (ttt--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    sort(a, a + n);
    bool good = false;
    s.clear();
    s.insert(0);
    for (int i = 0; i < n; ++i) {
      s2.clear();
      for (auto it = s.begin(); it != s.end(); ++it) s2.insert((*it) + a[i]);
      for (auto it = s2.begin(); it != s2.end(); ++it) {
        if (s.find(*it) != s.end())
          good = true;
        else
          s.insert(*it);
      }
    }
    if (good)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
set<int> s;
set<int> s2;
int main() {
  int ttt;
  cin >> ttt;
  while (ttt--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    sort(a, a + n);
    bool good = false;
    s.clear();
    s.insert(0);
    for (int i = 0; i < n; ++i) {
      s2.clear();
      for (auto it = s.begin(); it != s.end(); ++it) s2.insert((*it) + a[i]);
      for (auto it = s2.begin(); it != s2.end(); ++it) {
        if (s.find(*it) != s.end())
          good = true;
        else
          s.insert(*it);
      }
    }
    if (good)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

```