#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;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#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;
}

```