#include <bits/stdc++.h>
using namespace std;
const long long N = 1e4 + 10;
long long t, n;
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    long long l = -1 * n + 1;
    printf("%lld %lld\n", l, n);
  }
}

### Prompt

Your task is to create a Cpp solution 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;
const long long N = 1e4 + 10;
long long t, n;
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    long long l = -1 * n + 1;
    printf("%lld %lld\n", l, n);
  }
}

```