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 long long M = 1e9 + 7;
  const int N = 2e5 + 5;
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long mi(long long x) { return power(x, M - 2, M); }
  int nCr(int n, int r) {
    int mod = 1000000007;
    int C[r + 1];
    memset(C, 0, sizeof(C));
    C[0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = min(i, r); j > 0; j--) {
        C[j] = (C[j] + C[j - 1]) % mod;
      }
    }
    return C[r] % mod;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      int a[n];
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
      }
      map<long long, long long> mp;
      long long last = pow(2, n);
      int flag = 0;
      for (long long i = 0; i < last; i++) {
        long long sum = 0;
        for (long long j = 0; j < n; j++) {
          long long present = pow(2, j);
          long long check = present & i;
          if (check) {
            sum += a[j];
          }
        }
        mp[sum]++;
        if (mp[sum] > 1) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        cout << "YES" << endl;
      } else
        cout << "NO" << endl;
    }
    return 0;
  }
  