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;
  inline long long read() {
    long long s = 0, w = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') w = -1;
    for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48);
    return s * w;
  }
  inline void print(long long x, int op = 10) {
    if (!x) {
      putchar('0');
      if (op) putchar(op);
      return;
    }
    char F[40];
    long long tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
      F[cnt++] = tmp % 10 + '0';
      tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
    if (op) putchar(op);
  }
  inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
  long long qpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans *= a;
      b >>= 1;
      a *= a;
    }
    return ans;
  }
  long long qpow(long long a, long long b, long long mod) {
    long long ans = 1;
    while (b) {
      if (b & 1) (ans *= a) %= mod;
      b >>= 1;
      (a *= a) %= mod;
    }
    return ans % mod;
  }
  const int dir[][2] = {{0, 1}, {1, 0},  {0, -1}, {-1, 0},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
  const int MOD = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  const long long INF64 = 0x3f3f3f3f3f3f3f3f;
  const int N = 1e6 + 7;
  struct Node {
    long long val;
    int id;
    bool operator<(const Node& opt) const { return val < opt.val; }
  };
  long long n, m;
  set<int> st;
  int p[20];
  int solve() {
    n = read();
    int cnt = 0;
    for (int i = 1; i <= n; ++i) p[i] = read(), cnt += p[i] == 0;
    st.clear();
    st.insert(0);
    for (int i = 1; i <= n; ++i) {
      set<int> tmp = st;
      for (auto& it : st) {
        int x1 = it - p[i];
        int x2 = it + p[i];
        cnt += x1 == 0;
        cnt += x2 == 0;
        tmp.insert(x1), tmp.insert(x2);
      }
      st = tmp;
    }
    return cnt;
  }
  int main() {
    int T = read();
    for (int _ = 1; _ <= T; ++_) {
      cout << (solve() ? "YES" : "NO") << "\n";
    }
    return 0;
  }
  