#include <bits/stdc++.h>
using namespace std;
const long long int INF = 3e18;
const int inf = 1e9 + 10;
const int maxn = 200 + 10;
const int maxi = 4 * maxn;
const int mod = 1e9 + 7;
const int LOG = 25;
void solve() {
  int n;
  vector<int> vec;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    vec.push_back(a);
  }
  sort(vec.begin(), vec.end());
  bool flag = 0;
  for (int i : vec)
    if (i == 0) flag = 1;
  for (int i = 1; i < vec.size(); i++)
    if (vec[i] == vec[i - 1]) flag = 1;
  if (flag) {
    cout << "YES\n";
    return;
  }
  for (int mask = 0; mask < (1 << n); mask++) {
    long long int cnt = 0;
    for (int i = 0; i < vec.size(); i++)
      if (mask & (1 << i)) cnt += vec[i];
    for (int mask2 = 1; mask2 < (1 << n); mask2++) {
      if (mask + mask2 != (mask | mask2)) continue;
      long long int cnt2 = 0;
      for (int i = 0; i < vec.size(); i++)
        if ((1 << i) & mask2) cnt2 += vec[i];
      if (cnt2 == cnt) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  if (flag)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in CPP to 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 long long int INF = 3e18;
const int inf = 1e9 + 10;
const int maxn = 200 + 10;
const int maxi = 4 * maxn;
const int mod = 1e9 + 7;
const int LOG = 25;
void solve() {
  int n;
  vector<int> vec;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    vec.push_back(a);
  }
  sort(vec.begin(), vec.end());
  bool flag = 0;
  for (int i : vec)
    if (i == 0) flag = 1;
  for (int i = 1; i < vec.size(); i++)
    if (vec[i] == vec[i - 1]) flag = 1;
  if (flag) {
    cout << "YES\n";
    return;
  }
  for (int mask = 0; mask < (1 << n); mask++) {
    long long int cnt = 0;
    for (int i = 0; i < vec.size(); i++)
      if (mask & (1 << i)) cnt += vec[i];
    for (int mask2 = 1; mask2 < (1 << n); mask2++) {
      if (mask + mask2 != (mask | mask2)) continue;
      long long int cnt2 = 0;
      for (int i = 0; i < vec.size(); i++)
        if ((1 << i) & mask2) cnt2 += vec[i];
      if (cnt2 == cnt) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  if (flag)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```