description: Theofanis has a riddle for you and if you manage to solve it, he will give you a Cypriot snack halloumi for free (Cypriot cheese).
  
  You are given an integer n. You need to find two integers l and r such that -10^{18} ≤ l < r ≤ 10^{18} and l + (l + 1) + … + (r - 1) + r = n.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first and only line of each test case contains a single integer n (1 ≤ n ≤ 10^{18}).
  
  Output
  
  For each test case, print the two integers l and r such that -10^{18} ≤ l < r ≤ 10^{18} and l + (l + 1) + … + (r - 1) + r = n. 
  
  It can be proven that an answer always exists. If there are multiple answers, print any.
  
  Example
  
  Input
  
  
  7
  1
  2
  3
  6
  100
  25
  3000000000000
  
  
  Output
  
  
  0 1
  -1 2 
  1 2 
  1 3 
  18 22
  -2 7
  999999999999 1000000000001
  
  Note
  
  In the first test case, 0 + 1 = 1.
  
  In the second test case, (-1) + 0 + 1 + 2 = 2.
  
  In the fourth test case, 1 + 2 + 3 = 6.
  
  In the fifth test case, 18 + 19 + 20 + 21 + 22 = 100.
  
  In the sixth test case, (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 25.
solution: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _INPUT = """7
  # 1
  # 2
  # 3
  # 6
  # 100
  # 25
  # 3000000000000
  
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  def get_divisors(n):
      lower_divisors = []
      upper_divisors = []
      i = 1
      while i * i <= n:
          if n % i == 0:
              lower_divisors.append(i)
              if i != n // i:
                  upper_divisors.append(n//i)
          i += 1
      return lower_divisors + upper_divisors[::-1]
  
  
  def solve(N):
      # divs = get_divisors(2*N)
      # for d in divs:
      #     a = d
      #     b = (2*N)//d
      #     if (a+b-1) % 2 == 0 and (a-b+1) % 2 == 0:
      #         r = (a+b-1)//2
      #         l = (a-b+1)//2
      #         if -10**18 <= l < r <= 10**18:
      #             return [l, r]
      # return []
      return [-(N-1), N]
  
  T0 = int(input())
  for _ in range(T0):
      N = int(input())
      print(*solve(N))