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 maxn = 1e5 + 20;
inline int read() {
  int ans = 0;
  char last = ' ', ch = getchar();
  while (ch<'0' | ch> '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
int n;
int a[maxn];
map<int, int> ma1, ma2;
void solve() {
  n = read();
  int mn = 0x3f3f3f3f;
  ma1.clear(), ma2.clear();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i] < 0) a[i] = -a[i];
    ma1[a[i]]++;
  }
  bool ok = 0;
  if (n == 1 && a[1] == 0) ok = 1;
  for (auto it : ma1) {
    if (it.second >= 2) ok = 1;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (++ma2[abs(a[i] - a[j])] >= 2) ok = 1;
    }
  if (ok)
    puts("YES");
  else
    puts("NO");
}
int main() {
  int _t;
  _t = read();
  while (_t--) solve();
  return 0;
}
