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: import queue
  from math import inf, gcd, floor, sqrt, ceil, log, log2, log10, factorial
  import sys
  from collections import *
  from random import *
  from itertools import *
  
  sys.setrecursionlimit(99999)
  #sys.stdin = open("input.txt", 'r') # for debug purpose
  eps = sys.float_info.epsilon
  P = 2
  INF = 1e9 + 1
  MOD = 10 ** 9 + 7
  
  
  def is_prime(n):
      if n == 0 or n == 1:
          return False
      d = 2
      while d * d <= n:
          if n % d == 0:
              return False
          d += 1
      return True
  
  
  def div_up(n, k):
      if n % k == 0:
          return n // k
      else:
          return n // k + 1
  
  
  def num_len(n, base=10):
      if n == 0:
          return 1
      return int(floor(log(n, base) + 1))
  
  
  def dfs_sample(graph, cl, p, v):
      cl[v] = 1
      for to in graph[v]:
          if cl[to] == 1 and p[v] != to:
              # yay, cycle
              pass
          elif cl[to] == 0:
              p[to] = v
              dfs_sample(graph, cl, p, to)
      cl[v] = 2
  
  
  def down(a):
      for i in range(1, len(a)):
          if a[i] > a[i - 1]:
              return False
      return True
  
  
  def up(a):
      for i in range(1, len(a)):
          if a[i] < a[i - 1]:
              return False
      return True
  
  
  def code(c):
      return ord(c) - 32
  
  
  def _hash_(s):
      res, p = 0, 1
      for i in range(len(s)):
          res += (code(s[i]) * p)
          res %= MOD
          p *= P
          p %= MOD
      return res % MOD
  
  
  def remove_edge(v, u, graph):
      graph[v].remove(u)
      graph[u].remove(v)
  
  
  b = True
  
  
  def dfs_stack(start, graph, visited, flags):
      global b
      stack = []
      visited[start] = True
      stack.append(start)
      while len(stack) != 0:
          v = stack.pop()
          if not flags[v]:
              b = False
          for to in graph[v]:
              if not visited[to]:
                  visited[to] = True
                  stack.append(to)
  
  
  def all_eq(a):
      for i in range(len(a) - 1):
          if a[i] != a[i + 1]:
              return False
      return True
  
  
  def kmp(s, pr):
      n = len(s)
      pr[0] = 0
      for i in range(1, n):
          j = pr[i - 1]
          while j > 0 and s[i] != s[j]:
              j = pr[j - 1]
          if s[i] == s[j]:
              j += 1
          pr[i] = j
  
  
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  def gcd_advanced(a, b):
      if a == 0:
          return 0, 1, b
      x1, y1, g = gcd_advanced(b % a, a)
      return y1 - x1 * (b // a), x1, g
  
  
  def solve():
      n = int(input())
      print(-(n-1), n)
  
  
  for _ in range(int(input())):
      solve()
  
  
  def debug():
      pass
  
  # debug()