#include <bits/stdc++.h>
using namespace std;
long long binaryExponentiation(long long x, long long m) {
  if (x == 1) {
    return 0;
  } else {
    return 1 + (((binaryExponentiation((x / 2), m))));
  }
}
long long sumd(long long a) {
  long long ans = 0;
  while (a != 0) {
    ans += a % 10;
    a = a / 10;
  }
  return ans;
}
long long check(long long n, long long A[], long long m) {
  long long ans = 0;
  for (long long q = 0; q < m; q++) {
    if ((A[q] - n) > 0) ans += A[q] - n;
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  while (n--) {
    long long a;
    cin >> a;
    cout << -(a - 1) << " " << a << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long binaryExponentiation(long long x, long long m) {
  if (x == 1) {
    return 0;
  } else {
    return 1 + (((binaryExponentiation((x / 2), m))));
  }
}
long long sumd(long long a) {
  long long ans = 0;
  while (a != 0) {
    ans += a % 10;
    a = a / 10;
  }
  return ans;
}
long long check(long long n, long long A[], long long m) {
  long long ans = 0;
  for (long long q = 0; q < m; q++) {
    if ((A[q] - n) > 0) ans += A[q] - n;
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  while (n--) {
    long long a;
    cin >> a;
    cout << -(a - 1) << " " << a << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```