#include <bits/stdc++.h>
using namespace std;
int T, N;
bool ex[((int)1e6 + 10)];
int tv[10];
int v[10];
int reg[((int)1e6 + 10)];
int reginx = 0;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &N);
    bool dup = false;
    for (int i = 0; i < N; i++) {
      scanf("%d", v + i);
      if (v[i] == 0) dup = true;
      if (v[i] < 0) v[i] = -v[i];
    }
    sort(v, v + N);
    for (int i = 1; i < N && !dup; i++)
      if (v[i] == v[i - 1]) dup = true;
    if (dup) {
      printf("YES\n");
      continue;
    }
    bool ok = false;
    fill(ex, ex + ((int)1e6 + 10), false);
    reginx = 0;
    reg[reginx++] = v[0];
    ex[v[0]] = true;
    for (int i = 1; i < N && !ok; i++) {
      int val = v[i];
      int rsize = reginx;
      for (int k = 0; k < rsize; k++) {
        int newval = val + reg[k];
        if (ex[newval]) {
          ok = true;
          break;
        } else {
          reg[reginx++] = newval;
          ex[newval] = true;
        }
      }
      if (ex[val]) {
        ok = true;
        break;
      } else {
        reg[reginx++] = val;
        ex[val] = true;
      }
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Please formulate 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;
int T, N;
bool ex[((int)1e6 + 10)];
int tv[10];
int v[10];
int reg[((int)1e6 + 10)];
int reginx = 0;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &N);
    bool dup = false;
    for (int i = 0; i < N; i++) {
      scanf("%d", v + i);
      if (v[i] == 0) dup = true;
      if (v[i] < 0) v[i] = -v[i];
    }
    sort(v, v + N);
    for (int i = 1; i < N && !dup; i++)
      if (v[i] == v[i - 1]) dup = true;
    if (dup) {
      printf("YES\n");
      continue;
    }
    bool ok = false;
    fill(ex, ex + ((int)1e6 + 10), false);
    reginx = 0;
    reg[reginx++] = v[0];
    ex[v[0]] = true;
    for (int i = 1; i < N && !ok; i++) {
      int val = v[i];
      int rsize = reginx;
      for (int k = 0; k < rsize; k++) {
        int newval = val + reg[k];
        if (ex[newval]) {
          ok = true;
          break;
        } else {
          reg[reginx++] = newval;
          ex[newval] = true;
        }
      }
      if (ex[val]) {
        ok = true;
        break;
      } else {
        reg[reginx++] = val;
        ex[val] = true;
      }
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```