#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.1415926535897923846;
const long long int MOD = 1000000007;
const long long int N = 998244353;
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return res;
}
long long int modinverse(long long int a) { return power(a, MOD - 2); }
long long int n;
vector<long long int> a;
bool dfs(long long int index, long long int sum, bool used) {
  if (used && sum == 0) return true;
  if (index == n) return false;
  bool nas = false;
  nas = (nas || dfs(index + 1, sum, used));
  nas = (nas || dfs(index + 1, a[index] + sum, true));
  nas = (nas || dfs(index + 1, -a[index] + sum, true));
  return nas;
}
void solve() {
  cin >> n;
  a.clear();
  a.resize(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (dfs(0, 0, false))
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

### 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;
const long double PI = 3.1415926535897923846;
const long long int MOD = 1000000007;
const long long int N = 998244353;
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return res;
}
long long int modinverse(long long int a) { return power(a, MOD - 2); }
long long int n;
vector<long long int> a;
bool dfs(long long int index, long long int sum, bool used) {
  if (used && sum == 0) return true;
  if (index == n) return false;
  bool nas = false;
  nas = (nas || dfs(index + 1, sum, used));
  nas = (nas || dfs(index + 1, a[index] + sum, true));
  nas = (nas || dfs(index + 1, -a[index] + sum, true));
  return nas;
}
void solve() {
  cin >> n;
  a.clear();
  a.resize(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (dfs(0, 0, false))
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```