#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long fast(int x, int p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = (ret * x) % mod;
    x = (1LL * x * x) % mod;
    p >>= 1;
  }
  return ret;
}
void test() {
  long long n;
  scanf("%lld", &n);
  n *= 2;
  long long l, r;
  l = r = -1;
  for (long long i = 1; i <= n / i; i++) {
    if (n % i == 0) {
      long long x = i;
      long long y = n / i;
      l = x - y + 1;
      r = x + y - 1;
      if (l % 2 == 0 && r % 2 == 0) {
        l >>= 1LL;
        r >>= 1LL;
        break;
      }
    }
    if (l != -1) break;
  }
  if (l > r) swap(l, r);
  printf("%lld %lld\n", l, r);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) test();
}

### Prompt

In CPP, your task is to solve 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 int mod = 1e9 + 7;
long long fast(int x, int p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = (ret * x) % mod;
    x = (1LL * x * x) % mod;
    p >>= 1;
  }
  return ret;
}
void test() {
  long long n;
  scanf("%lld", &n);
  n *= 2;
  long long l, r;
  l = r = -1;
  for (long long i = 1; i <= n / i; i++) {
    if (n % i == 0) {
      long long x = i;
      long long y = n / i;
      l = x - y + 1;
      r = x + y - 1;
      if (l % 2 == 0 && r % 2 == 0) {
        l >>= 1LL;
        r >>= 1LL;
        break;
      }
    }
    if (l != -1) break;
  }
  if (l > r) swap(l, r);
  printf("%lld %lld\n", l, r);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) test();
}

```