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: t = int(input())
  mp = [1, -1]
  for _ in range(t):
      n = int(input())
      a=list(map(int,input().split()))
      if n % 2 == 0:
  
          print(' '.join(str(a[i + j] * mp[j]) for i in range(0, n, 2) for j in (1, 0)))
          continue
      y = sorted(map(abs, a))
      for i in range(n-1):
          if y[i] == y[i + 1]:
              break
      else:
          print(' '.join(str(a[i + j] * mp[j]) for i in range(0, n - 3, 2) for j in (1, 0)),
                a[-1] * a[-2], a[-1] * a[-3], -2 * a[-2] * a[-3])
          continue
      v = y[i]
      p1, p2, p3 = -1, -1, -1
      for j in range(n):
          if abs(a[j]) == v:
              if p1 == -1:
                  p1 = j
              elif p2 == -1:
                  p2 = j
              else:
                  break
  
      if 1000000000 - sum(y) < 10000:
          for i in range(n - 2):
              if y[i] == y[i + 1] == y[i + 2]:
                  break
          v = y[i]
          p1, p2, p3 = -1, -1, -1
          for j in range(n):
              if abs(a[j]) == v:
                  if p1 == -1:
                      p1 = j
                  elif p2 == -1:
                      p2 = j
                  elif p3 == -1:
                      p3 = j
                  else:
                      break
  
          vs = [0] * n
          op = -1
          for i in range(n):
              if i == p1 or i == p2 or i == p3:
                  continue
              if op == -1:
                  op = i
              else:
                  vs[op] = a[i]
                  vs[i] = -a[op]
                  op = -1
          vs[p1], vs[p2], vs[p3] = 1, 1, -2
          if a[p1] < 0:
              vs[p1] = -vs[p1]
          if a[p2] < 0:
              vs[p2] = -vs[p2]
          if a[p3] < 0:
              vs[p3] = -vs[p3]
          print(' '.join(str(e) for e in vs))
          
      else:
          vs = [0] * n
          op = -1
          for i in range(n):
              if i == p1 or i == p2:
                  continue
              if op == -1:
                  op = i
              else:
                  vs[op] = a[i]
                  vs[i] = -a[op]
                  op = -1
          vs[p1] = a[op]
          vs[p2] = a[op]
          vs[op] = -2 * a[p1]
          if (a[p1] < 0) != (a[p2] < 0):
              vs[p2] = -vs[p2]
          
          print(' '.join(str(e) for e in vs))
  