#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vvi = vector<vector<int>>;
bool get_next(vector<int> &u) {
  int i = u.size() - 1;
  while (i >= 0 && u[i] == 1) {
    --i;
  }
  if (i < 0) {
    return false;
  }
  ++u[i];
  for (int j = i + 1; j < (int)u.size(); ++j) {
    u[j] = -1;
  }
  return true;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &e : a) {
    cin >> e;
    e = abs(e);
  }
  sort(begin(a), end(a));
  if (find(begin(a), end(a), 0) != end(a)) {
    cout << "YES\n";
    return;
  }
  if (n == 1) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1]) {
      cout << "YES\n";
      return;
    }
  }
  vector<int> u(n, -1);
  do {
    int cnt_nonzero = 0;
    int s = 0;
    for (int i = 0; i < n; ++i) {
      s += u[i] * a[i];
      cnt_nonzero += (u[i] != 0);
    }
    if (cnt_nonzero > 0 && s == 0) {
      cout << "YES\n";
      return;
    }
  } while (get_next(u));
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a CPP solution 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;
using ll = long long;
using vvi = vector<vector<int>>;
bool get_next(vector<int> &u) {
  int i = u.size() - 1;
  while (i >= 0 && u[i] == 1) {
    --i;
  }
  if (i < 0) {
    return false;
  }
  ++u[i];
  for (int j = i + 1; j < (int)u.size(); ++j) {
    u[j] = -1;
  }
  return true;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &e : a) {
    cin >> e;
    e = abs(e);
  }
  sort(begin(a), end(a));
  if (find(begin(a), end(a), 0) != end(a)) {
    cout << "YES\n";
    return;
  }
  if (n == 1) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1]) {
      cout << "YES\n";
      return;
    }
  }
  vector<int> u(n, -1);
  do {
    int cnt_nonzero = 0;
    int s = 0;
    for (int i = 0; i < n; ++i) {
      s += u[i] * a[i];
      cnt_nonzero += (u[i] != 0);
    }
    if (cnt_nonzero > 0 && s == 0) {
      cout << "YES\n";
      return;
    }
  } while (get_next(u));
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```