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;
  using ll = long long;
  using P = pair<ll, ll>;
  using vec = vector<ll>;
  using vecp = vector<P>;
  using mat = vector<vec>;
  using matp = vector<vecp>;
  const ll MOD = 998244353;
  const ll INF = 1e18;
  ll Pow(ll a, ll n) {
    if (n == 0) return 1;
    if (n == 1) return a;
    if (n % 2 == 1) return (a * Pow(a, n - 1));
    ll t = Pow(a, n / 2);
    return (t * t);
  }
  int main() {
    ll T;
    cin >> T;
    for (long long l = 0; l < (T); ++l) {
      ll N;
      cin >> N;
      vec A(N);
      for (long long i = 0; i < (N); ++i) cin >> A.at(i);
      if (N == 1) {
        if (A.at(0) == 0) {
          cout << "YES" << endl;
        } else {
          cout << "NO" << endl;
        }
        continue;
      }
      ll t = 0;
      for (long long i = 0; i < (N); ++i) {
        for (long long tmp = 0; tmp < (Pow(3, N - 1)); ++tmp) {
          ll k = 0, s = tmp;
          for (long long j = 0; j < (N); ++j) {
            if (j == i) {
              continue;
            }
            if (s % 3 == 1)
              k += A.at(j);
            else if (s % 3 == 2)
              k -= A.at(j);
            s /= 3;
          }
          if (k == A.at(i)) t = 1;
        }
      }
      if (t == 1) {
        cout << "YES" << endl;
      } else {
        cout << "NO" << endl;
      }
    }
  }
  