description: 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.
solution: #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;
  }
  