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;
  const double EPS = 1e-9;
  const long long INF = 2e9, MOD = 1e9 + 7;
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      for (long long i = 0; i < n; i++) cin >> a[i];
      for (long long ind = 0; ind < n; ind++) {
        vector<long long> tmp = a;
        tmp.erase(tmp.begin() + ind);
        if (a[ind] == 0) {
          goto l;
        }
        for (long long i = 0; i < (1 << (n - 1)); i++) {
          long long s = 0;
          vector<long long> left;
          for (long long j = 0; j < n - 1; j++) {
            if (i & (1 << j)) {
              s += tmp[j];
            } else {
              left.push_back(tmp[j]);
            }
          }
          long long cur;
          long long num = (long long)left.size();
          for (long long j = 0; j < (1 << num); j++) {
            cur = 0;
            for (long long x = 0; x < num; x++) {
              if (j & (1 << x)) {
                cur += left[x];
                if (abs(cur - s) == abs(a[ind]) || abs(cur + s) == abs(a[ind])) {
                  goto l;
                }
              }
            }
          }
        }
      }
      cout << "NO"
           << "\n";
      continue;
    l:
      cout << "YES"
           << "\n";
    }
    return 0;
  }
  