#include <bits/stdc++.h>
using namespace std;
namespace DBH {
const int DIR4[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int DIR8[8][2] = {{1, 0},   {-1, 0}, {0, 1},  {0, -1},
                        {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
const int MOD197 = 1e9 + 7;
const int MOD19 = 1e9;
const int MOD353 = 998244353;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const pair<int, int> NILP = {0, -1};
inline int LT(int x) { return x << 1; }
inline int RT(int x) { return (x << 1) | 1; }
inline int LB(int x) { return x & -x; }
inline int MSK(int x) { return 1 << x; }
inline int HF(int x) { return x >> 1; }
inline int DB(int x) { return x << 1; }
inline int GT(int x, int i) { return (x >> i) & 1; }
inline int ST(int x, int i) { return x | (1 << i); }
inline int ODD(int x) { return x & 1; }
inline int SGN(int x) { return x >= 0 ? (x == 0 ? 0 : 1) : -1; }
template <typename T>
inline void qread(T &X) {
  X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
}  // namespace DBH
using namespace DBH;
void solve() {
  long long t;
  cin >> t;
  cout << -t + 1 << " " << t << endl;
}
int main() {
  ;
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) 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;
namespace DBH {
const int DIR4[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int DIR8[8][2] = {{1, 0},   {-1, 0}, {0, 1},  {0, -1},
                        {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
const int MOD197 = 1e9 + 7;
const int MOD19 = 1e9;
const int MOD353 = 998244353;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const pair<int, int> NILP = {0, -1};
inline int LT(int x) { return x << 1; }
inline int RT(int x) { return (x << 1) | 1; }
inline int LB(int x) { return x & -x; }
inline int MSK(int x) { return 1 << x; }
inline int HF(int x) { return x >> 1; }
inline int DB(int x) { return x << 1; }
inline int GT(int x, int i) { return (x >> i) & 1; }
inline int ST(int x, int i) { return x | (1 << i); }
inline int ODD(int x) { return x & 1; }
inline int SGN(int x) { return x >= 0 ? (x == 0 ? 0 : 1) : -1; }
template <typename T>
inline void qread(T &X) {
  X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
}  // namespace DBH
using namespace DBH;
void solve() {
  long long t;
  cin >> t;
  cout << -t + 1 << " " << t << endl;
}
int main() {
  ;
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```