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: import sys
  import math
  import heapq
  import bisect
  from collections import Counter
  from collections import defaultdict
  from io import BytesIO, IOBase
  from itertools import permutations
  import string
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          import os
          self.os = os
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
          self.BUFSIZE = 8192
  
      def read(self):
          while True:
              b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              self.os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  
  def get_int():
      return int(input())
  
  
  def get_ints():
      return list(map(int, input().split(' ')))
  
  
  def get_int_grid(n):
      return [get_ints() for _ in range(n)]
  
  
  def get_str():
      return input().strip()
  
  
  def get_strs():
      return get_str().split(' ')
  
  
  def flat_list(arr):
      return [item for subarr in arr for item in subarr]
  
  
  def yes_no(b):
      if b:
          return "YES"
      else:
          return "NO"
  
  
  def binary_search(good, left, right, delta=1, right_true=False):
      """
      Performs binary search
      ----------
      Parameters
      ----------
      :param good: Function used to perform the binary search
      :param left: Starting value of left limit
      :param right: Starting value of the right limit
      :param delta: Margin of error, defaults value of 1 for integer binary search
      :param right_true: Boolean, for whether the right limit is the true invariant
      :return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
              alternatively returns False if no such value found
      """
  
      limits = [left, right]
      while limits[1] - limits[0] > delta:
          if delta == 1:
              mid = sum(limits) // 2
          else:
              mid = sum(limits) / 2
          if good(mid):
              limits[int(right_true)] = mid
          else:
              limits[int(~right_true)] = mid
      if good(limits[int(right_true)]):
          return limits[int(right_true)]
      else:
          return False
  
  
  def prefix_sums(a):
      p = [0]
      for x in a:
          p.append(p[-1] + x)
      return p
  
  
  
  def solve_d():
      n = get_int()
      a = get_ints()
      ans = [None] * n
  
      C = defaultdict(list)
      for i, x in enumerate(a):
          C[x].append(i)
          if len(C[x]) == 2:
              i, j = C[x].pop(), C[x].pop()
              ans[i] = 1
              ans[j] = -1
              last_i, last_j = i, j
      rem = []
      for k, v in C.items():
          while v:
              rem.append((k, v.pop()))
  
  
      if len(rem) == 1:
          k, v = rem.pop()
          ans[v] = 2 * a[last_i] * a[last_j]
          ans[last_i] = - a[v] * a[last_j]
          ans[last_j] = - a[v] * a[last_i]
      if len(rem) % 2 == 1:
          k1, v1 = rem.pop()
          k2, v2 = rem.pop()
          k3, v3 = rem.pop()
          ans[v1] = 2 * a[v2] * a[v3]
          ans[v2] = - a[v1] * a[v3]
          ans[v3] = - a[v1] * a[v2]
      while rem:
          k1, v1 = rem.pop()
          k2, v2 = rem.pop()
          ans[v1] = a[v2]
          ans[v2] = -a[v1]
  
      return ans
  
  
  t = get_int()
  for _ in range(t):
      print(*solve_d())
  