#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("abm,fma,mmx,avx,popcnt,avx2,sse,sse2,sse3")
using namespace std;
void Seeyouagain() {
  int_fast64_t n;
  cin >> n;
  if (n % 2)
    cout << n / 2 << " " << (n + 1) / 2;
  else {
    cout << -(n - 1) << " " << n;
  }
}
signed main() {
  ios_base::sync_with_stdio(!cin.tie(nullptr));
  int_fast64_t testc = 1;
  cin >> testc;
  while (testc--) {
    Seeyouagain();
    cout << "\n";
  }
  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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("abm,fma,mmx,avx,popcnt,avx2,sse,sse2,sse3")
using namespace std;
void Seeyouagain() {
  int_fast64_t n;
  cin >> n;
  if (n % 2)
    cout << n / 2 << " " << (n + 1) / 2;
  else {
    cout << -(n - 1) << " " << n;
  }
}
signed main() {
  ios_base::sync_with_stdio(!cin.tie(nullptr));
  int_fast64_t testc = 1;
  cin >> testc;
  while (testc--) {
    Seeyouagain();
    cout << "\n";
  }
  return 0;
}

```