description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long int gcd(long long int n1, long long int n2) {
    if (n2 != 0)
      return gcd(n2, n1 % n2);
    else
      return n1;
  }
  struct boom {
    int arr[5];
    int ind;
  };
  vector<vector<long long int> > master;
  void gen(long long int i, vector<long long int> vec) {
    if (i == (10)) {
      master.push_back(vec);
      return;
    }
    vec.push_back(0);
    gen(i + 1, vec);
    vec[vec.size() - 1] = 1;
    gen(i + 1, vec);
    vec[vec.size() - 1] = 2;
    gen(i + 1, vec);
  }
  void solve() {
    long long int i, j, n;
    cin >> n;
    long long int arr[n];
    set<long long int> set1;
    vector<long long int> vec;
    for (i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i] < 0) arr[i] = -arr[i];
      if (set1.find(arr[i]) == set1.end()) vec.push_back(arr[i]);
      set1.insert(arr[i]);
    }
    for (i = 0; i < n; i++) {
      if (arr[i] == 0) {
        cout << "YES" << endl;
        return;
      }
    }
    if (n == 1) {
      if (arr[0] == 0) {
        cout << "YES" << endl;
        return;
      } else {
        cout << "NO" << endl;
        return;
      }
    }
    if ((n > (set1.size())) || (set1.size() == 1)) {
      cout << "YES" << endl;
      return;
    }
    if (n == 2) {
      if (set1.size() == 2) {
        cout << "NO" << endl;
        return;
      } else {
        cout << "YES" << endl;
        return;
      }
    }
    for (i = 0; i < (master.size()); i++) {
      for (long long int fin = 0; fin < (vec.size()); fin++) {
        long long int k = 0;
        for (j = 0; j < (vec.size()); j++) {
          if (j == fin) continue;
          if (master[i][j] == 1)
            k += vec[j];
          else if (master[i][j] == 2)
            k -= vec[j];
        }
        if (k == vec[fin]) {
          cout << "YES" << endl;
          return;
        }
      }
    }
    cout << "NO" << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    vector<long long int> emp;
    gen(0, emp);
    long long int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  