description: Vupsen and Pupsen were gifted an integer array. Since Vupsen doesn't like the number 0, he threw away all numbers equal to 0 from the array. As a result, he got an array a of length n.
  
  Pupsen, on the contrary, likes the number 0 and he got upset when he saw the array without zeroes. To cheer Pupsen up, Vupsen decided to come up with another array b of length n such that ∑_{i=1}^{n}a_i ⋅ b_i=0. Since Vupsen doesn't like number 0, the array b must not contain numbers equal to 0. Also, the numbers in that array must not be huge, so the sum of their absolute values cannot exceed 10^9. Please help Vupsen to find any such array b!
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2 ⋅ t lines contain the description of test cases. The description of each test case consists of two lines.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the length of the array.
  
  The second line contains n integers a_1, a_2, …, a_n (-10^4 ≤ a_i ≤ 10^4, a_i ≠ 0) — the elements of the array a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print n integers b_1, b_2, …, b_n — elements of the array b (|b_1|+|b_2|+… +|b_n| ≤ 10^9, b_i ≠ 0, ∑_{i=1}^{n}a_i ⋅ b_i=0).
  
  It can be shown that the answer always exists.
  
  Example
  
  Input
  
  
  3
  2
  5 5
  5
  5 -2 10 -9 4
  7
  1 2 3 4 5 6 7
  
  
  Output
  
  
  1 -1
  -1 5 1 -1 -1
  -10 2 2 -3 5 -1 -1
  
  Note
  
  In the first test case, 5 ⋅ 1 + 5 ⋅ (-1)=5-5=0. You could also print 3 -3, for example, since 5 ⋅ 3 + 5 ⋅ (-3)=15-15=0
  
  In the second test case, 5 ⋅ (-1) + (-2) ⋅ 5 + 10 ⋅ 1 + (-9) ⋅ (-1) + 4 ⋅ (-1)=-5-10+10+9-4=0.
solution: from sys import stdin
  input = stdin.readline
  
  for _ in range(int(input())):
      n = int(input())
      a = [int(x) for x in input().split()]
      res = [0] * n
      d, dd = dict(), dict()
      s = set(range(n))
      chk = []
  
      for i, v in enumerate(a):
          try:
              d[v] += 1
              dd[v].append(i)
          except:
              d[v] = 1
              dd[v] = [i]
  
      for x in d.keys():
          if d[x] <= 1:
              continue
  
          if d[x] % 2:
              for y in range(0, d[x] - 1, 2):
                  res[dd[x][y]] = 1
                  res[dd[x][y + 1]] = -1
                  s.remove(dd[x][y])
                  s.remove(dd[x][y + 1])
          else:
              for y in range(0, d[x], 2):
                  res[dd[x][y]] = 1
                  res[dd[x][y + 1]] = -1
                  s.remove(dd[x][y])
                  s.remove(dd[x][y + 1])
  
          if len(chk) == 0:
              chk += [dd[x][0], dd[x][1]]
  
      if len(s) == 0:
          print(*res)
          continue
  
      if len(s) % 2 == 0:
          s = list(s)
  
          for x in range(0, len(s), 2):
              res[s[x]] = a[s[x + 1]]
              res[s[x + 1]] = -a[s[x]]
  
          print(*res)
          continue
  
      if len(s) == 1:
          s = list(s)
          
          t1, t2, t3 = s[0], chk[0], chk[1]
          res[t1] = a[t2] * a[t3]
          res[t2] = a[t1] * a[t3]
          res[t3] = -2 * a[t1] * a[t2]
  
          print(*res)
          continue
  
      if len(s) == 3:
          s = list(s)
  
          res[s[0]] = a[s[1]] * a[s[2]]
          res[s[1]] = a[s[0]] * a[s[2]]
          res[s[2]] = -2 * a[s[0]] * a[s[1]]
  
          print(*res)
          continue
      
      s = list(s)
      res[s[0]] = a[s[1]] * a[s[2]]
      res[s[1]] = a[s[0]] * a[s[2]]
      res[s[2]] = -2 * a[s[0]] * a[s[1]]
  
      for x in range(3, len(s), 2):
          res[s[x]] = a[s[x + 1]]
          res[s[x + 1]] = -a[s[x]]
  
      print(*res)