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
  import math
  
  if sys.subversion[0] == "PyPy":
      import io, atexit
  
      sys.stdout = io.BytesIO()
      atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
  
      sys.stdin = io.BytesIO(sys.stdin.read())
      input = lambda: sys.stdin.readline().rstrip()
  
  t = int(raw_input())
  
  # python codesforces210710.py < input.txt
  for x in range(t):
      n = int(raw_input())
      str1 = raw_input()
      li = str1.split(' ')
      li = map(int, li)
      tf = 0
      ssu = sum(li)
      vec = [-1] * n
      matri = []
      newvec = vec[:]
      matri.append(newvec[:])
      co = 0
      for i in range(1, 3 ** n):
          if newvec[0] != 1:
              newvec[0] += 1
              co = 0
          else:
              newvec[0] = -1
              co = 1
          for j in range(1, n):
              if newvec[j] == 1 and co == 1:
                  newvec[j] = -1
                  co = 1
              else:
                  newvec[j] = newvec[j] + co
                  co = 0
          matri.append(newvec[:])
          # print(matri)
  
  
          ssu = [a * b for a, b in zip(matri[i], li)]
          if sum(ssu) == 0:
              # print(matri[i])
              tf += 1
          if tf == 2:
              # print(matri[i])
              break
      if tf > 1:
          print('YES')
      else:
          print('NO')
  
  