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;
bool dfs(string s, string t, long long ind, long long node) {
  if (ind == t.length() - 1) {
    return true;
  }
  if (node == s.length() - 1) {
    return false;
  }
  for (long long j = 1; j < s.length() - node; j += 2) {
    if (s[node + j] == t[ind + 1] && dfs(s, t, ind + 1, node + j)) {
      return true;
    }
  }
  return false;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v;
  map<long long, long long> mp;
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    if (mp[abs(x)] == 0) {
      mp[abs(x)]++;
      cnt++;
    }
    v.push_back(x);
  }
  long long val = (n * (n - 1)) / 2;
  if (cnt < val) {
    cout << "YES" << endl;
    return;
  }
  if (n == 1) {
    if (v[0] == 0) {
      cout << "YES" << endl;
      return;
    }
    cout << "NO" << endl;
    return;
  }
  if (n == 2) {
    if (abs(v[0]) == abs(v[1])) {
      cout << "YES" << endl;
      return;
    }
    cout << "NO" << endl;
    return;
  }
  if (abs(v[0] - v[1]) == abs(v[2])) {
    cout << "YES" << endl;
    return;
  }
  if (abs(v[2] - v[1]) == abs(v[0])) {
    cout << "YES" << endl;
    return;
  }
  if (abs(v[0] - v[2]) == abs(v[1])) {
    cout << "YES" << endl;
    return;
  }
  cout << "NO" << endl;
  return;
}
int main() {
  cin.sync_with_stdio(false), cin.tie(NULL);
  long long t;
  cin >> t;
  long long x = 0;
  while (t--) {
    x++;
    solve();
  }
  return 0;
}
