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, os
  from io import BytesIO, IOBase
  from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
  from collections import defaultdict as dd, deque
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from bisect import bisect_left as bl, bisect_right as br, bisect
  
  # region fastio
  BUFSIZE = 8192
   
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
   
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) 
  
  mod = pow(10, 9) + 7
  mod2 = 998244353
          
  def inp(): return stdin.readline().strip()
  def iinp(): return int(inp())
  def out(var, end="\n"): stdout.write(str(var)+"\n")
  def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
  def lmp(): return list(mp())
  def mp(): return map(int, inp().split())
  def l1d(n, val=0): return [val for i in range(n)]
  def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
  def ceil(a, b): return (a+b-1)//b
  S1 = 'abcdefghijklmnopqrstuvwxyz'
  S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  def isprime(x):
      if x<=1: return False
      if x in (2, 3): return True
      if x%2 == 0: return False
      for i in range(3, int(sqrt(x))+1, 2):
          if x%i == 0: return False
      return True
   
  for _ in range(iinp()):
      n = iinp()
      arr = lmp()
      ansl = []
      md = {}
      flg = False
      for i in range(1<<n):
          sm = 0
          for j in range(n):
              if i&(1<<j):
                  sm += arr[j]
          if sm in md:
              flg = True
              break
          md[sm] = 1
      print("YES" if flg else "NO")