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 i64 = long long;
  using db = double;
  using u32 = unsigned int;
  using u64 = unsigned long long;
  using db = double;
  using pii = pair<int, int>;
  using vi = vector<int>;
  using qi = queue<int>;
  using pqi = priority_queue<int>;
  using si = set<int>;
  const int INF = 0x3f3f3f3f;
  const ll INFLL = 0x3f3f3f3f3f3f3f3f;
  const double PI = acos(-1.0);
  template <class T>
  inline bool chkmin(T& a, T b) {
    return b < a ? a = b, true : false;
  }
  template <class T>
  inline bool chkmax(T& a, T b) {
    return a < b ? a = b, true : false;
  }
  int _w, _t;
  FILE* _f;
  const int N = 1024;
  int n, a[N], mx, b[N];
  void solve() {
    memset(b, 0, sizeof b);
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      a[i] = abs(a[i]);
    }
    mx = 1 << n;
    for (int i = 0; i < mx; i++) {
      for (int j = 0; j < n; j++) {
        if ((i >> j) & 1) {
          b[i] += a[j];
        }
      }
    }
    for (int i = 0; i < mx; i++) {
      for (int j = i + 1; j < mx; j++) {
        if (b[i] == b[j]) {
          printf("YES\n");
          return;
        }
      }
    }
    printf("NO\n");
  }
  int main() {
    _w = scanf("%d", &_t);
    while (_t--) solve();
    return 0;
  }
  