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 <typename T>
  istream& operator>>(istream& is, vector<T>& v) {
    for (T& x : v) is >> x;
    return is;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    for (int i = 0; i < (int)v.size(); i++) {
      os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
    }
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, const map<T, U>& m) {
    os << '{';
    for (auto itr = m.begin(); itr != m.end();) {
      os << '(' << itr->first << ',' << itr->second << ')';
      if (++itr != m.end()) os << ',';
    }
    os << '}';
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, const unordered_map<T, U>& m) {
    os << '{';
    for (auto itr = m.begin(); itr != m.end();) {
      os << '(' << itr->first << ',' << itr->second << ')';
      if (++itr != m.end()) os << ',';
    }
    os << '}';
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const set<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
      os << *itr;
      if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const multiset<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
      os << *itr;
      if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const unordered_set<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
      os << *itr;
      if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const deque<T>& v) {
    for (int i = 0; i < (int)v.size(); i++) {
      os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
    }
    return os;
  }
  template <int i, typename T>
  void print_tuple(ostream&, const T&) {}
  template <int i, typename T, typename H, class... Args>
  void print_tuple(ostream& os, const T& t) {
    if (i) os << ',';
    os << get<i>(t);
    print_tuple<i + 1, T, Args...>(os, t);
  }
  template <typename... Args>
  ostream& operator<<(ostream& os, const tuple<Args...>& t) {
    os << '{';
    print_tuple<0, tuple<Args...>, Args...>(os, t);
    return os << '}';
  }
  void debug_out() { cerr << '\n'; }
  template <class Head, class... Tail>
  void debug_out(Head&& head, Tail&&... tail) {
    cerr << head;
    if (sizeof...(Tail) > 0) cerr << ", ";
    debug_out(move(tail)...);
  }
  template <typename T>
  T gcd(T x, T y) {
    return y != 0 ? gcd(y, x % y) : x;
  }
  template <typename T>
  T lcm(T x, T y) {
    return x / gcd(x, y) * y;
  }
  int topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
  int topbit(long long t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
  int botbit(signed a) { return a == 0 ? 32 : __builtin_ctz(a); }
  int botbit(long long a) { return a == 0 ? 64 : __builtin_ctzll(a); }
  int popcount(signed t) { return __builtin_popcount(t); }
  int popcount(long long t) { return __builtin_popcountll(t); }
  bool ispow2(int i) { return i && (i & -i) == i; }
  template <class T>
  T ceil(T x, T y) {
    assert(y >= 1);
    return (x > 0 ? (x + y - 1) / y : x / y);
  }
  template <class T>
  T floor(T x, T y) {
    assert(y >= 1);
    return (x > 0 ? x / y : (x - y + 1) / y);
  }
  template <class T1, class T2>
  inline bool chmin(T1& a, T2 b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T1, class T2>
  inline bool chmax(T1& a, T2 b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  const int INF = 1e9;
  const long long IINF = 1e18;
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  const char dir[4] = {'D', 'R', 'U', 'L'};
  const long long MOD = 1000000007;
  void solve() {
    long long n;
    cin >> n;
    long long l = -(n - 1), r = n;
    cout << l << ' ' << r << '\n';
  }
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    for (; t--;) solve();
    return 0;
  }
  