#include <bits/stdc++.h>
using namespace std;
void SMG() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << -(n - 1) << " " << n << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  SMG();
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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;
void SMG() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << -(n - 1) << " " << n << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  SMG();
  return 0;
}

```