#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e2, N1 = maxn + 5, p = 1000000007;
template <typename _T>
void read(_T &ret) {
  ret = 0;
  _T fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  ret = ret * fh;
}
int T, n;
int a[N1], b[N1];
int check() {
  for (int s = 0; s < (1 << n); s++) {
    for (int t = s;; t = (t - 1) & s) {
      int now = 0;
      for (int i = 0; i < n; i++) {
        if (s & (1 << i)) {
          if (t & (1 << i))
            now += a[i];
          else
            now -= a[i];
        }
      }
      for (int i = 0; i < n; i++) {
        if (!(s & (1 << i)))
          if (abs(now) == a[i]) return 1;
      }
      if (!t) break;
    }
  }
  return 0;
}
void solve() {
  read(n);
  for (int i = 0; i < n; i++) read(a[i]), a[i] = abs(a[i]);
  int fl = check();
  if (fl)
    puts("YES");
  else
    puts("NO");
}
int main() {
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 1e2, N1 = maxn + 5, p = 1000000007;
template <typename _T>
void read(_T &ret) {
  ret = 0;
  _T fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  ret = ret * fh;
}
int T, n;
int a[N1], b[N1];
int check() {
  for (int s = 0; s < (1 << n); s++) {
    for (int t = s;; t = (t - 1) & s) {
      int now = 0;
      for (int i = 0; i < n; i++) {
        if (s & (1 << i)) {
          if (t & (1 << i))
            now += a[i];
          else
            now -= a[i];
        }
      }
      for (int i = 0; i < n; i++) {
        if (!(s & (1 << i)))
          if (abs(now) == a[i]) return 1;
      }
      if (!t) break;
    }
  }
  return 0;
}
void solve() {
  read(n);
  for (int i = 0; i < n; i++) read(a[i]), a[i] = abs(a[i]);
  int fl = check();
  if (fl)
    puts("YES");
  else
    puts("NO");
}
int main() {
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}

```