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.
#include <bits/stdc++.h>
using namespace std;
const int int_inf = 0x7fffffff;
const long long ll_inf = (1ll << 63) - 1;
const double eps = 1e-7;
inline int read() {
  int f = 1, x = 0;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  x = ch - '0';
  while (isdigit(ch = getchar())) x = x * 10 + ch - '0';
  return x * f;
}
inline void read(int num, ...) {
  va_list res;
  va_start(res, &num);
  for (int i = 1; i <= num; i++) *va_arg(res, int*) = read();
  va_end(res);
  return;
}
const int N = 20;
int n, a[N], b[N], ok = 0;
inline void dfs(int pos) {
  if (ok) return;
  if (pos > n) {
    for (int i = 1; i <= n; i++) {
      int fl = 0;
      for (int j = 1; j <= n; j++)
        for (int k = 1; k <= n; k++)
          if (a[i] == b[j] - b[k]) fl = 1;
      if (!fl) return;
    }
    ok = 1;
    return;
  }
  for (int i = 1; i < pos; i++) {
    b[pos] = b[i] + a[pos];
    if (b[pos] == b[pos - 1]) ok = 1;
    dfs(pos + 1);
  }
  return;
}
inline void work() {
  cin >> n;
  ok = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  b[1] = 0;
  dfs(2);
  if (ok || a[1] == 0)
    cout << "YES\n";
  else
    cout << "NO\n";
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) work();
  return 0;
}
