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 int maxn = 1e2, N1 = maxn + 5, p = 1000000007;
  template <typename _T>
  void read(_T &ret) {
    ret = 0;
    _T fh = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') fh = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      ret = ret * 10 + c - '0';
      c = getchar();
    }
    ret = ret * fh;
  }
  int T, n;
  int a[N1], b[N1];
  int check() {
    for (int s = 0; s < (1 << n); s++) {
      for (int t = s;; t = (t - 1) & s) {
        int now = 0;
        for (int i = 0; i < n; i++) {
          if (s & (1 << i)) {
            if (t & (1 << i))
              now += a[i];
            else
              now -= a[i];
          }
        }
        for (int i = 0; i < n; i++) {
          if (!(s & (1 << i)))
            if (abs(now) == a[i]) return 1;
        }
        if (!t) break;
      }
    }
    return 0;
  }
  void solve() {
    read(n);
    for (int i = 0; i < n; i++) read(a[i]), a[i] = abs(a[i]);
    int fl = check();
    if (fl)
      puts("YES");
    else
      puts("NO");
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      solve();
    }
    return 0;
  }
  