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;
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  pair<int, int> operator*(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator%(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first % x.first, p.second % x.second);
    return np;
  }
  pair<long long, long long> operator*(const pair<long long, long long> &p,
                                       const pair<int, int> &x) {
    pair<long long, long long> np =
        make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator*(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first * x.first, p.second * x.second);
    return np;
  }
  pair<long long, long long> operator+(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first + x.first, p.second + x.second);
    return np;
  }
  pair<long long, long long> operator-(const pair<long long, long long> &p,
                                       const pair<long long, long long> &x) {
    pair<long long, long long> np =
        make_pair(p.first - x.first, p.second - x.second);
    return np;
  }
  pair<int, int> operator+(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first + x.first, p.second + x.second);
    return np;
  }
  pair<int, int> operator-(const pair<int, int> &p, const pair<int, int> &x) {
    pair<int, int> np = make_pair(p.first - x.first, p.second - x.second);
    return np;
  }
  vector<int> prime;
  bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.first > b.first);
  }
  bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
    return a.second > b.second;
  }
  bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.second < b.second);
  }
  long long powerm(long long a, long long k) {
    if (!k) {
      return 1;
    }
    long long b = powerm(a, k / 2);
    b = b * b % 998244353;
    if (k % 2) {
      return a * b % 998244353;
    } else {
      return b;
    }
  }
  long long power(long long a, long long b) {
    if (b == 1) return a;
    if (b == 0) return 1;
    long long m1 = power(a, b / 2);
    if (b % 2) return m1 * m1 * a;
    return m1 * m1;
  }
  void PFactor(int x) {
    prime.clear();
    for (int i = 2; i <= x / i; i++)
      if (x % i == 0) {
        while (x % i == 0) x /= i, prime.push_back(i);
      }
    if (x > 1) prime.push_back(x);
  }
  bool isprime(long long a) {
    if (a <= 1) return false;
    if (a == 2 || a == 3) return true;
    if (a % 2 == 0 || a % 3 == 0) return false;
    for (long long i = 5; i * i <= a; i = i + 6) {
      if (a % i == 0 || a % (i + 2) == 0) return false;
    }
    return true;
  }
  long long read() {
    long long a = 0;
    char b = 1, c;
    do
      if ((c = getchar()) == 45) b = -1;
    while (c < 48 || c > 57);
    do a = (a << 3) + (a << 1) + (c & 15);
    while ((c = getchar()) > 47 && c < 58);
    return a * b;
  }
  void write(long long x, char c) {
    if (x < 0) putchar(45), x = -x;
    char a[20], s = 0;
    do a[++s] = x % 10 | 48;
    while (x /= 10);
    do putchar(a[s]);
    while (--s);
    putchar(c);
  }
  const int N = 2e6 + 20;
  vector<int> g[N];
  int n = 0, m = 0, k = 0;
  int a = 0, b = 0, c = 0;
  string str;
  int d[N] = {0};
  int ans[N];
  bool flag = false;
  void dfs(int deep) {
    if (flag) return;
    if (deep == n + 1) return;
    int x = d[deep];
    for (int i = 0; i <= deep - 1; i++) {
      ans[deep] = ans[i] + x;
      for (int j = 0; j <= deep - 1; j++)
        if (ans[deep] == ans[j]) flag = true;
      dfs(deep + 1);
    }
  }
  int main() {
    int T;
    cin >> T;
    while (T--) {
      cin >> n;
      for (int i = 1; i <= n; i++) d[i] = read();
      flag = false;
      for (int i = 0; i <= n + 10; i++) ans[i] = 0;
      dfs(1);
      if (flag)
        cout << "YES\n";
      else
        cout << "NO\n";
    }
  }
  