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 sys
  input = lambda : sys.stdin.readline().rstrip()
  
  sys.setrecursionlimit(2*10**5+10)
  write = lambda x: sys.stdout.write(x+"\n")
  debug = lambda x: sys.stderr.write(x+"\n")
  writef = lambda x: print("{:.12f}".format(x))
  
  
  t = int(input())
  ans = []
  for _ in range(t):
      n = int(input())
      a = list(map(int, input().split()))
      if 0 in a:
          res = 1
      else:
          for b in range(3**n):
              l0 = []
              l1 = []
              for i in range(n):
                  b,v = divmod(b, 3)
                  if v==1:
                      l0.append(a[i])
                  elif v==2:
                      l1.append(a[i])
              if not l0:
                  continue
              if sum(l0)==sum(l1, 0):
                  res = 1
                  break
          else:
              res = 0
      if res:
          ans.append("Yes")
      else:
          ans.append("No")
  write("\n".join(ans))