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: #include <bits/stdc++.h>
  using namespace std;
  double EPS = 1e-9;
  int INF = 1000000005;
  long long INFF = 1000000000000000005LL;
  double PI = acos(-1);
  int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
  int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
  int INFH = 2e9 + 10;
  int binarySearch(int arr[], int l, int r, int x) {
    if (r >= l) {
      int mid = l + (r - l) / 2;
      if (arr[mid] == x) return mid;
      if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
      return binarySearch(arr, mid + 1, r, x);
    }
    return -1;
  }
  int binexp(long long a, long long b) {
    if (b == 0) return 1;
    long long res = binexp(a, b / 2);
    if (b & 1)
      return a * res * res;
    else
      return res * res;
  }
  int factorial(int i) { return i ? factorial(i - 1) * i : 1; }
  bool prime(long long n) {
    if (n <= 2) return false;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) return false;
    }
    return true;
  }
  int di(long long n) {
    long long sum = 0;
    while (n > 0) {
      n /= 10;
      sum++;
    }
    return sum;
  }
  long long M = 1e9 + 7;
  int smallest_divisor(int n) {
    int i;
    for (i = 2; i <= sqrt(n); ++i) {
      if (n % i == 0) {
        return i;
      }
    }
    return n;
  }
  void solved() {
    long long n;
    cin >> n;
    cout << -(n - 1) << " " << n << endl;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solved();
    }
  }
  