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: '''
      Auther: ghoshashis545 Ashis Ghosh
      College: Jalpaiguri Govt Enggineering College
  
  '''
  from os import path
  from io import BytesIO, IOBase
  import sys
  from heapq import heappush,heappop
  from functools import cmp_to_key as ctk
  from collections import deque,Counter,defaultdict as dd 
  from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  from itertools import permutations
  from datetime import datetime
  from math import ceil,sqrt,log,gcd
  def ii():return int(input())
  def si():return input().rstrip()
  def mi():return map(int,input().split())
  def li():return list(mi())
  abc='abcdefghijklmnopqrstuvwxyz'
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0],[0,0,1,-1]
  
  def bo(i):
      return ord(i)-ord('0')
  
  file = 1
  def ceil(a,b):
      return (a+b-1)//b
  
  
  
  def isprime(n):
      for i in range(2,int(sqrt(n))+1):
          if n%i==0:
              return False
      return True
  
  
  def solve():
  
      
      for _ in range(1,ii()+1):
          n = ii()
          a = li()
          res = [0]*n
          for i in range(n//2):
              l,r = a[i],a[n-i-1]
              p = (l*r)//gcd(l,r)
              res[i] = (p//l)
              res[n-i-1] = (-p)//r
          
          if n&1:
              r1 = n//2
              tot = sum(res)
              ok = False
              for i in range(n):
                  if i==r1:
                      continue
                  l,r = a[i],a[r1]
                  p = (l*r)//gcd(l,r)
                  x1 = p//l
                  x2 = -(p//r)
                  if res[i]+x1:
                      res[i] += x1
                      res[r1]+=x2
                      ok = True
                      break
              if ok==False:
                  res=[0]*n
                  l1,r1 = 0,n-2
                  for i in range(n//2):
                      l,r = a[l1],a[r1]
                      p = (l*r)//gcd(l,r)
                      res[l1] = (p//l)
                      res[r1] = (-p)//r
                      r1-=1
                      l1+=1
                  r1 = n-1
                  tot = sum(res)
                  for i in range(n):
                      if i==r1:
                          continue
                      l,r = a[i],a[r1]
                      p = (l*r)//gcd(l,r)
                      x1 = p//l
                      x2 = -(p//r)
                      if res[i]+x1:
                          res[i] += x1
                          res[r1]+=x2
                          ok = True
                          break
  
          print(*res) 
          s = 0
          # for i in range(n):
          #     s += (res[i]*a[i])
          # print(s)
         
  
        
          
          
  
  
  
  
  
  
  
              
  
  
  
  
  
      
          
  
  
  
  
  
  
  
  
          
  if __name__ =="__main__":
  
      if(file):
      
         if path.exists('input.txt'):
             sys.stdin=open('input.txt', 'r')
             sys.stdout=open('output.txt','w')
         else:
             input=sys.stdin.readline
      solve()
  