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: 
  /**
   * Write a description of class bengue here.
   *
   * @author (your name)
   * @version (a version number or a date)
   */
  import java.util.*;
  import java.io.*;
  public class bengue
  {
      public static void main(String[] args)
      {
          Scanner in = new Scanner(System.in);
          int num = in.nextInt();
          for(int i = 0; i < num; i++)
          {
              int n = in.nextInt();
              int[] a = new int[n];
              for(int j = 0; j < n; j++)
              {
                  a[j] = Math.abs(in.nextInt());
              }
              Arrays.sort(a);
              boolean b = false;
              for(int j = 1; j < Math.pow(2, n) && !b; j++)
              {
                  boolean[] r = new boolean[n];
                  int counter2 = 0;
                  int lol = j;
                  for(int k = 0; k < n; k++)
                  {
                      int l = (int)(lol / Math.pow(2, n-k-1));
                      r[k] = l != 0;
                      
                      if(r[k])
                      {
                          counter2++;
                          lol -= Math.pow(2, n-k-1);
                      }
                  }
                  int[] h = new int[counter2];
                  int counter3 = 0;
                  
                  for(int k = 0; k < n; k++)
                  {
                      if(r[k])
                      {
                          h[counter3] = a[k];
                          counter3++;
                      }
                  }
                  for(int k = 0; k < Math.pow(2, counter2) && !b; k++)
                  {
                      int lol2 = k;
                      int sum = 0;
                      for(int m = counter2 - 1; m >= 0 && !b; m--)
                      {
                          if((int)(lol2 / Math.pow(2, m)) == 0)
                              sum -= h[m];
                          else
                          {
                              sum += h[m];
                              lol2 -= Math.pow(2, m);
                          }
                          
                      }
                      for(int m = 0; m < n && !b; m++)
                      {
                          if(!r[m] && a[m] == Math.abs(sum))
                              b = true;
                      }
                  }
              }
              for(int j = 0; j < n-1 && !b; j++)
              {
                  if(a[j] == a[j+1])
                      b = true;
              }
              for(int j = 0; j < n && !b; j++)
              {
                  b = a[j] == 0;
              }
              if(b)
                  System.out.println("YES");
              else
                  System.out.println("NO");
          }
          
      }
  }
  