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;
  int power(int x, unsigned int y) {
    int res = 1;
    while (y > 0) {
      if (y & 1) res = res * x;
      y = y >> 1;
      x = x * x;
    }
    return res;
  }
  long long binExpo(long long a, long long n) {
    long long res = 1;
    while (n > 0) {
      if (n % 2) {
        res *= a, n--;
      } else {
        a *= a;
        n /= 2;
      }
    }
    return res;
  }
  int modInverse(int a, int m) {
    int m0 = m;
    int y = 0, x = 1;
    if (m == 1) return 0;
    while (a > 1) {
      int q = a / m;
      int t = m;
      m = a % m, a = t;
      t = y;
      y = x - q * y;
      x = t;
    }
    if (x < 0) x += m0;
    return x;
  }
  long long fac(long long n) {
    long long res = 1;
    for (long long i = 2; i <= n; i++) {
      res *= i;
    }
    return res;
  }
  long long nCr(long long n, long long r) {
    return fac(n) / (fac(r) * fac(n - r));
  }
  void solve() {
    long long n;
    cin >> n;
    if (n & 1) {
      cout << n / 2 << " " << n / 2 + 1 << endl;
    } else {
      cout << 1 - n << " " << n << endl;
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  