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, math, cmath, time, collections
  from collections import deque, Counter, OrderedDict, defaultdict
  from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
  from math import ceil, floor, log, log2, sqrt, gcd, factorial, pow, pi
  from bisect import bisect_left, bisect_right
  
  start_time = time.time()
  
  
  def solve(a, n):
      """
      The actual solution begins here
      c = a + b
      print(c)
      """
      cur = [0]
      for i in range(n):
          cur = cur + [x + a[i] for x in cur] + [x - a[i] for x in cur]
          # print(f"The current is: {cur}")
      print('YES' if sum(x == 0 for x in cur) > 1 else 'NO')
  
  def main():
      """
      Main function dedicated to get the I/P
      a, b = map(int, input().split())
      solve(a, b)
      """
      n = int(input())
      arr = list(map(int, input().split()))
      solve(arr, n)
  
  
  if __name__ == "__main__":
      testcases = int(input())
      for i in range(testcases):
          main()