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: def find_common(a1, a2):
      list_len = len(a1)
      list_len2 = len(a2)
      a3 = []
      iz = jz = 0
      while iz < list_len and jz < list_len2:
          if a1[iz] < a2[jz]:
              iz += 1
          elif a2[jz] < a1[iz]:
              jz += 1
          else:
              a3.append(a1[iz])    
              iz +=1
              jz +=1
      return a3
  
  t=int(input())
  for i in range(t):
      y=0
      n=int(input())
      a=list(map(int,input().split()))
      if 0 in a:
          print("YES")
      else:
          if n<=5:
              for j in range(1,3**n):
                  tz=0
                  for k in range(n):
                      if (j//3**k)%3==1:
                          tz+=a[k]
                      elif (j//3**k)%3==2:
                          tz-=a[k]
                  if tz==0:
                      y=1
                      break
              if y==0:
                  print("NO")
              else:
                  print("YES")
          else:
              l=[]
              for j in range(1,3**5):
                  tz=0
                  for k in range(5):
                      if (j//3**k)%3==1:
                          tz+=a[k]
                      elif (j//3**k)%3==2:
                          tz-=a[k]
                  l.append(abs(tz))
              l=list(set(l))
              l.sort()
              a=a[5:]
              ln=[]
              for j in range(1,3**len(a)):
                  tz=0
                  for k in range(len(a)):
                      if (j//3**k)%3==1:
                          tz+=a[k]
                      elif (j//3**k)%3==2:
                          tz-=a[k]
                  ln.append(tz)
              ln=list(set(ln))
              ln.sort()
              if find_common(l,ln):
                  print("YES")
              elif 0 in ln:
                  print("YES")
              elif 0 in l:
                  print("YES")
              else:
                  print("NO")