#include <bits/stdc++.h>
#pragma GCC diagnostic push
#pragma GCC diagnostic ignored "-Wunused-variable"
#pragma GCC diagnostic ignored "-Wc++11-extensions"
#pragma GCC diagnostic ignored "-Wunused-const-variable"
using namespace std;
int max(int a, int b) { return (a > b) ? a : b; }
int min(int a, int b) { return (a < b) ? a : b; }
double max(double a, double b) { return (a > b) ? a : b; }
double min(double a, double b) { return (a < b) ? a : b; }
void ain(vector<int>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
int Ceil(double d) { return ((int)d == d) ? d : d + 1; }
int gcd(int a, int b) { return (a == 0) ? b : gcd(b % a, a); }
int n;
vector<int> a;
void f(int ind, int sum, vector<int>& v, set<int>& all) {
  all.insert(sum);
  if (ind == v.size()) return;
  f(ind + 1, sum + v[ind], v, all);
  f(ind + 1, sum - v[ind], v, all);
  f(ind + 1, sum, v, all);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int _;
  cin >> _;
  while (_--) {
    cin >> n;
    a.assign(n, 0);
    ain(a);
    bool done = 0;
    for (int i = 0; i < n; i++) {
      vector<int> v;
      for (int j = 0; j < n; j++)
        if (i != j) v.push_back(a[j]);
      set<int> sub;
      f(0, 0, v, sub);
      if (sub.find(a[i]) != sub.end()) {
        done = 1;
        cout << "YES\n";
        break;
      }
    }
    if (!done) cout << "NO\n";
  }
  return 0;
}
#pragma GCC diagnostic pop

### Prompt

Please create a solution in cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC diagnostic push
#pragma GCC diagnostic ignored "-Wunused-variable"
#pragma GCC diagnostic ignored "-Wc++11-extensions"
#pragma GCC diagnostic ignored "-Wunused-const-variable"
using namespace std;
int max(int a, int b) { return (a > b) ? a : b; }
int min(int a, int b) { return (a < b) ? a : b; }
double max(double a, double b) { return (a > b) ? a : b; }
double min(double a, double b) { return (a < b) ? a : b; }
void ain(vector<int>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
int Ceil(double d) { return ((int)d == d) ? d : d + 1; }
int gcd(int a, int b) { return (a == 0) ? b : gcd(b % a, a); }
int n;
vector<int> a;
void f(int ind, int sum, vector<int>& v, set<int>& all) {
  all.insert(sum);
  if (ind == v.size()) return;
  f(ind + 1, sum + v[ind], v, all);
  f(ind + 1, sum - v[ind], v, all);
  f(ind + 1, sum, v, all);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int _;
  cin >> _;
  while (_--) {
    cin >> n;
    a.assign(n, 0);
    ain(a);
    bool done = 0;
    for (int i = 0; i < n; i++) {
      vector<int> v;
      for (int j = 0; j < n; j++)
        if (i != j) v.push_back(a[j]);
      set<int> sub;
      f(0, 0, v, sub);
      if (sub.find(a[i]) != sub.end()) {
        done = 1;
        cout << "YES\n";
        break;
      }
    }
    if (!done) cout << "NO\n";
  }
  return 0;
}
#pragma GCC diagnostic pop

```