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;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  void _print(int x) { cerr << x; }
  void _print(long x) { cerr << x; }
  void _print(long long x) { cerr << x; }
  void _print(unsigned x) { cerr << x; }
  void _print(unsigned long x) { cerr << x; }
  void _print(unsigned long long x) { cerr << x; }
  void _print(float x) { cerr << x; }
  void _print(double x) { cerr << x; }
  void _print(long double x) { cerr << x; }
  void _print(char x) { cerr << '\'' << x << '\''; }
  void _print(const char* x) { cerr << '\"' << x << '\"'; }
  void _print(const string& x) { cerr << '\"' << x << '\"'; }
  void _print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename A, typename B>
  void _print(const pair<A, B>& p);
  template <typename A, typename B, typename C>
  void _print(const tuple<A, B, C>& tp);
  template <typename A, typename B, typename C, typename D>
  void _print(const tuple<A, B, C, D>& tp);
  template <typename A>
  void _print(const A& x);
  template <class A, class B, class C>
  istream& operator>>(istream& in, tuple<A, B, C>& tp) {
    return in >> get<0>(tp) >> get<1>(tp) >> get<2>(tp);
  }
  template <class A, class B>
  istream& operator>>(istream& in, pair<A, B>& p) {
    return in >> p.first >> p.second;
  }
  template <class A>
  istream& operator>>(istream& in, vector<A>& x) {
    for (A& i : x) in >> i;
    return in;
  }
  template <typename A, typename B>
  void _print(const pair<A, B>& p) {
    cerr << "(";
    _print(p.first);
    cerr << ", ";
    _print(p.second);
    cerr << ')';
  }
  template <typename A, typename B, typename C>
  void _print(const tuple<A, B, C>& tp) {
    cerr << "(";
    _print(get<0>(tp));
    cerr << ", ";
    _print(get<1>(tp));
    cerr << ", ";
    _print(get<2>(tp));
    cerr << ')';
  }
  template <typename A, typename B, typename C, typename D>
  void _print(const tuple<A, B, C, D>& tp) {
    cerr << "(";
    _print(get<0>(tp));
    cerr << ", ";
    _print(get<1>(tp));
    cerr << ", ";
    _print(get<2>(tp));
    cerr << ", ";
    _print(get<3>(tp));
    cerr << ')';
  }
  template <typename A>
  void _print(const A& x) {
    int f = 0;
    cerr << '{';
    for (auto i : x) cerr << (f++ ? ", " : ""), _print(i);
    cerr << "}";
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << " ";
    _print(H);
    dbg_out(T...);
  }
  const int MOD = 1e9 + 7;
  const char nl = '\n';
  const long long INF = 1e18;
  const int MX = 2e5 + 5;
  void solve() {
    long long n;
    cin >> n;
    cout << -(n - 1) << ' ' << n << nl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tc = 1;
    cin >> tc;
    while (tc--) solve();
    return 0;
  }
  