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 INF = 0x7fffffff;
const int N = 1e5 + 5;
inline int read() {
  int x = 0, rev = 0, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') rev = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return rev ? -x : x;
}
int n, a[N];
bool dfs(int i, int step, int sum) {
  if (sum == a[i]) return 1;
  if (step == n + 1) return 0;
  int flg = 0;
  if (i == step) return dfs(i, step + 1, sum);
  flg |= dfs(i, step + 1, sum + a[step]) + dfs(i, step + 1, sum - a[step]) +
         dfs(i, step + 1, sum);
  return flg;
}
int main() {
  int T = read();
  while (T--) {
    int flg = 0;
    n = read();
    for (int i = 0; i < n; i++) {
      a[i] = read();
    }
    for (int i = 0; i < n; i++) {
      flg |= dfs(i, 1, 0);
    }
    puts(flg ? "YES" : "NO");
  }
}
