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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  const long long MOD = 1000000007;
  const long double PI = 3.141592653589793238;
  const long long pi = 31415926;
  const long long inf = 1000000000000000000;
  const long long small_inf = INT_MAX;
  const int N = 300002;
  int power(int x, int n) {
    int res = 1;
    while (n > 0) {
      if (n & 1) res = res * x;
      x = x * x;
      n >>= 1;
    }
    return res;
  }
  int add(int x, int y) { return (x + y) % MOD; }
  int mult(int x, int y) { return (x * y) % MOD; }
  void init() {}
  void solve() {
    int n;
    cin >> n;
    std::vector<int> v(n, 0);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    for (int i = 0; i < power(2, n); i++) {
      std::vector<int> v1;
      if (i == 0) continue;
      for (int j = 0; j < n; j++) {
        if ((i & (1 << j))) v1.push_back(v[j]);
      }
      {
        int n = v1.size();
        for (int i = 0; i < power(2, n); i++) {
          int x = 0;
          for (int j = 0; j < n; j++) {
            if (i & (1 << j)) {
              x += v1[j];
            } else
              x -= v1[j];
          }
          if (x == 0) {
            cout << "YES" << endl;
            return;
          }
        }
      }
    }
    cout << "NO\n";
  }
  void clear_global() {}
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    int t23 = 1, tt23 = 1;
    cin >> t23;
    init();
    while (tt23 <= t23) {
      solve();
      tt23++;
      clear_global();
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
         << "ms\n";
  }
  