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: import java.io.*;
  import java.util.*;
  
  public class MainClass {
      public static void main(String[] args) {
          Reader in = new Reader(System.in);
          StringBuilder stringBuilder = new StringBuilder();
          int t = in.nextInt();
          while (t-- > 0) {
              int n = in.nextInt();
              long[] a = new long[n];
              for (int i = 0; i < n; i++) {
                  a[i] = in.nextLong();
              }
              boolean ff = false;
              for (int i = 1; i < (1 << n); i++) {
                  for (int j = 0; j < (1 << solve(i)); j++) {
                      int sum = 0;
                      int count = 0;
                      for (int bit = 0; bit < n; bit++) {
                          int mask = 1 << bit;
                          int sign = 1;
                          if ((j & (1 << count)) > 0) {
                              sign = -1;
                          }
                          if ((i & mask) > 0) {
                              sum += sign * a[bit];
                              count++;
                          }
                      }
                      if (sum == 0) {
                          ff = true;
                      }
                  }
              }
              if (ff) {
                  stringBuilder.append("YES\n");
              } else {
                  stringBuilder.append("NO\n");
              }
          }
          System.out.println(stringBuilder);
      }
      public static int solve(int x) {
          int count = 0;
          while (x > 0) {
              x = x & (x - 1);
              count++;
          }
          return count;
      }
  }
  class Reader {
      public BufferedReader reader;
      public StringTokenizer tokenizer;
  
      public Reader(InputStream stream) {
          reader = new BufferedReader(new InputStreamReader(stream), 32768);
          tokenizer = null;
      }
  
      public String next() {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
              try {
                  tokenizer = new StringTokenizer(reader.readLine());
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
          }
          return tokenizer.nextToken();
      }
  
      public int nextInt() {
          return Integer.parseInt(next());
      }
  
      public long nextLong() {
          return Long.parseLong(next());
      }
  
  }
  /*
  If atleast one zero found, then b = a => YES
  Else if a_i - a_j = a_k, then use b_k = 0 => YES
  Else if duplicates in a, then YES
  
  1 2 5
  
   */