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;
vector<vector<long long>> adj;
map<long, bool> vis, viss;
vector<long long> parent, sz;
long long components = 0;
long long dr[] = {1, 0, -1, 0};
long long dc[] = {0, 1, 0, -1};
int spf[1000000 + 1];
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int gcd(int a, int b, int& x, int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int x1, y1;
  int d = gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(nullptr);
  long long t, temp;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      cin >> temp;
      v.push_back(temp);
    }
    string ans = "NO";
    for (long long i = 0; i < n; i++) {
      if (v[i] == 0) ans = "YES";
      for (long long j = 0; j < n - 1; j++) {
        for (long long k = j + 1; k < n; k++) {
          if (i != j && i != k) {
            if (v[i] == v[j] || v[i] == v[k] || v[i] == v[j] + v[k] ||
                v[i] == abs(v[j] - v[k]))
              ans = "YES";
          }
          if (abs(v[j] - v[k]) >= 0 && abs(v[j] - v[k]) < n) ans = "YES";
          if (abs(v[j] + v[k]) >= 0 && abs(v[j] + v[k]) < n) ans = "YES";
        }
      }
    }
    if (n == 1 && v[0] == 0) ans = "YES";
    cout << ans << "\n";
  }
  return 0;
}
