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;
  template <typename type_first, typename type_second>
  ostream& operator<<(ostream& out, const pair<type_first, type_second> P) {
    out << '{' << P.first << ',' << P.second << '}';
    return out;
  }
  template <typename type_iterator>
  string ite2string(type_iterator begin, type_iterator end) {
    stringstream out;
    bool inserted = false;
    out << '[';
    while (begin != end) {
      out << *begin << ", ";
      inserted = true;
      begin++;
    }
    string ans = out.str();
    if (inserted) {
      ans.pop_back();
      ans.pop_back();
    }
    return ans + ']';
  }
  template <class stream, class iterable,
            class begin = decltype(begin(declval<iterable>()))>
  stream& operator<<(stream& out, const iterable I) {
    out << ite2string(I.begin(), I.end());
    return out;
  }
  bool constructor(unsigned n, unsigned l, vector<int>& sign, vector<int>& a) {
    if (n == l) {
      int p = 0;
      unsigned c0 = 0;
      for (unsigned int i = 0; i < n; i++) {
        if (sign[i] == 0) {
          c0++;
        }
        p += sign[i] * a[i];
      }
      if (c0 != n and p == 0) {
        ;
        return true;
      } else {
        return false;
      }
    }
    sign[l] = +1;
    if (constructor(n, l + 1, sign, a)) {
      return true;
    }
    sign[l] = 0;
    if (constructor(n, l + 1, sign, a)) {
      return true;
    }
    sign[l] = -1;
    if (constructor(n, l + 1, sign, a)) {
      return true;
    }
    return false;
  }
  bool check(vector<int>& a) {
    unsigned int n = a.size();
    sort((a).begin(), (a).end());
    for (unsigned int i = 0; i < n; i++) {
      if (a[i] == 0) {
        return true;
      }
      if (i != n - 1) {
        if (a[i] == a[i + 1]) {
          return true;
        }
      }
    }
    vector<int> sign(n, 0);
    return constructor(n, 0, sign, a);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    unsigned t;
    cin >> t;
    while (t--) {
      unsigned int n;
      cin >> n;
      vector<int> a(n);
      for (auto& i : a) {
        cin >> i;
      }
      if (check(a)) {
        cout << "YES" << '\n';
      } else {
        cout << "NO" << '\n';
      }
    }
    return 0;
  }
  