#include <bits/stdc++.h>
using namespace std;
const long long int mod = (long long int)(998244353);
const long long int mod2 = (long long int)1e9 + 7;
long long int test = 1;
vector<long long int> read(int n) {
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  return v;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  long long int i, j, k, m, n, cnt = 0, ans = 0, sum = 0;
  cin >> n;
  if (n & 1LL) {
    cout << n / 2 << " " << n / 2 + (n % 2) << "\n";
  } else {
    cout << -(n - 1) << " " << n << "\n";
  }
  test++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i = 1;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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 int mod = (long long int)(998244353);
const long long int mod2 = (long long int)1e9 + 7;
long long int test = 1;
vector<long long int> read(int n) {
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  return v;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  long long int i, j, k, m, n, cnt = 0, ans = 0, sum = 0;
  cin >> n;
  if (n & 1LL) {
    cout << n / 2 << " " << n / 2 + (n % 2) << "\n";
  } else {
    cout << -(n - 1) << " " << n << "\n";
  }
  test++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i = 1;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```