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;
  void sol(long long int x, long long int y, long long int ind,
           long long int &flag, vector<long long int> &a) {
    if (x == y && x != 0) {
      flag = 1;
      return;
    }
    if (ind == a.size()) return;
    sol(x + a[ind], y, ind + 1, flag, a);
    if (flag == 1) return;
    sol(x, y, ind + 1, flag, a);
    if (flag == 1) return;
    sol(x, y + a[ind], ind + 1, flag, a);
    if (flag == 1) return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long int t, l, flag, od, ev, sum, ma, mi, i, j, k, n, m, ans, tot, count,
        x, y;
    cin >> t;
    while (t--) {
      cin >> n;
      vector<long long int> a;
      for (i = 0; i < n; i++) {
        cin >> x;
        a.push_back(abs(x));
      }
      sort(a.begin(), a.end());
      if (a[0] == 0)
        cout << "YES\n";
      else if (n == 1)
        cout << "NO\n";
      else if (n == 2) {
        if (a[0] == a[1])
          cout << "YES\n";
        else
          cout << "NO\n";
      } else {
        flag = 0;
        for (i = 0; i < n - 1; i++) {
          if (a[i] == a[i + 1]) {
            flag = 1;
            break;
          }
        }
        if (flag == 1)
          cout << "YES\n";
        else {
          for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
              for (k = j + 1; k < n; k++) {
                if (i != j && i != k) {
                  if (a[j] + a[k] == a[i] || a[k] - a[j] == a[i]) {
                    flag = 1;
                    break;
                  }
                }
              }
              if (flag == 1) break;
            }
            if (flag == 1) break;
          }
          if (flag == 1)
            cout << "YES\n";
          else {
            sol(0, 0, 0, flag, a);
            if (flag == 1)
              cout << "YES\n";
            else
              cout << "NO\n";
          }
        }
      }
    }
    return 0;
  }
  