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: # Problem: D. Вупсень, Пупсень и 0
  # Contest: Codeforces - Codeforces Round #750 (Div. 2)
  # URL: https://codeforces.com/contest/1582/problem/D
  # Memory Limit: 256 MB
  # Time Limit: 1000 ms
  # 
  # Powered by CP Editor (https://cpeditor.org)
  
  import bisect
  
  mod = 10 ** 9 + 7
  eps = 10 ** -9
  
  
  def __gcd(a, b):
  	return a if b == 0 else __gcd(b, a % b)
  	
  
  def __lcm(a, b):
  	return a * b / __gcd(a, b)
  
  
  def __fact(n):
  	return 1 if n == 1 else n * __fact(n - 1)
  
  
  def __mex(a):
  	mex = 0
  	a.sort()
  	for x in a:
  		if x <= mex:
  			mex += 1
  		else:
  			break
  	return mex
  
  def __dist(x1, y1, x2, y2):
  	return (x1 - x2) ** 2 + (y1 - y2) ** 2
  
  def __getprimes(n):
      isprime = [True for i in range(n + 1)]
      primes = []
       
      for i in range(2, n + 1):
          if isprime[i]:
              for j in range(i * i, n + 1, i):
                  isprime[j] = False
      for i in range(2, n + 1):
          if isprime[i]:
              primes.append(i)
      return primes
      
  def __cntprimediv(n):
  	ret = 0
  	x = n
  	i = 2
  	while i * i <= x:
  		while n % i == 0:
  			n //= i
  			ret += 1
  		i += 1
  	if n > 1:
  		ret += 1
  	return ret
  	
  def __primefactors(n):
  	ret = []
  	x = n
  	i = 2
  	while i * i <= x:
  		while n % i == 0:
  			ret.append(i)
  			n //= i
  		i += 1
  	if n > 1:
  		ret.append(n)
  	return ret
  	
  	
  def __sumdigit(n):
  	ret = 0
  	while n > 0:
  		ret += n % 10
  		n //= 10
  	return ret
  	
  	
  def solve(t):
  	n = int(input())
  	a = list(map(int, input().split()))
  	b = [0 for i in range(n)]
  	k = n
  	if n % 2 > 0:
  		k = n - 3
  	for i in range(0, k, 2):
  		gcd = __gcd(abs(a[i]), abs(a[i + 1]))
  		sign = 1
  		if a[i] * a[i + 1] > 0:
  			sign = -1
  		b[i] = abs(a[i + 1]) // gcd
  		b[i + 1] = sign * abs(a[i]) // gcd
  	if k < n:
  		if a[-3] + a[-2] != 0:
  			gcd = __gcd(abs(a[-1]), abs(a[-3] + a[-2]))
  			sign = 1
  			if a[-1] * (a[-2] + a[-3]) > 0:
  				sign = -1
  			b[-3] = abs(a[-1]) // gcd
  			b[-2] = abs(a[-1]) // gcd
  			b[-1] = sign * abs(a[-2] + a[-3]) // gcd
  			
  		elif a[-2] + a[-1] != 0:
  			gcd = __gcd(abs(a[-3]), abs(a[-1] + a[-2]))
  			sign = 1
  			if a[-3] * (a[-2] + a[-1]) > 0:
  				sign = -1
  			b[-1] = abs(a[-3]) // gcd
  			b[-2] = abs(a[-3]) // gcd
  			b[-3] = sign * abs(a[-2] + a[-1]) // gcd
  		else:
  			
  			gcd = __gcd(abs(a[-2]), abs(a[-3] + a[-2]))
  			sign = 1
  			if a[-2] * (a[-2] + a[-3]) > 0:
  				sign = -1
  			b[-1] = abs(a[-2]) // gcd
  			b[-3] = abs(a[-2]) // gcd
  			b[-2] = sign * abs(a[-3] + a[-1]) // gcd
  	
  	print(*b)
  
  t = 1
  t = int(input())
  
  
  for i in range(t):
  	solve(i + 1)
  