#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const int N = 2e5 + 5;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mi(long long x) { return power(x, M - 2, M); }
int nCr(int n, int r) {
  int mod = 1000000007;
  int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) {
      C[j] = (C[j] + C[j - 1]) % mod;
    }
  }
  return C[r] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    map<long long, long long> mp;
    long long last = pow(2, n);
    int flag = 0;
    for (long long i = 0; i < last; i++) {
      long long sum = 0;
      for (long long j = 0; j < n; j++) {
        long long present = pow(2, j);
        long long check = present & i;
        if (check) {
          sum += a[j];
        }
      }
      mp[sum]++;
      if (mp[sum] > 1) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      cout << "YES" << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 M = 1e9 + 7;
const int N = 2e5 + 5;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mi(long long x) { return power(x, M - 2, M); }
int nCr(int n, int r) {
  int mod = 1000000007;
  int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) {
      C[j] = (C[j] + C[j - 1]) % mod;
    }
  }
  return C[r] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    map<long long, long long> mp;
    long long last = pow(2, n);
    int flag = 0;
    for (long long i = 0; i < last; i++) {
      long long sum = 0;
      for (long long j = 0; j < n; j++) {
        long long present = pow(2, j);
        long long check = present & i;
        if (check) {
          sum += a[j];
        }
      }
      mp[sum]++;
      if (mp[sum] > 1) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      cout << "YES" << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}

```