Problem: Ayrat is looking for the perfect code. He decided to start his search from an infinite field tiled by hexagons. For convenience the coordinate system is introduced, take a look at the picture to see how the coordinates of hexagon are defined: 

<image> <image> Ayrat is searching through the field. He started at point (0, 0) and is moving along the spiral (see second picture). Sometimes he forgets where he is now. Help Ayrat determine his location after n moves.

Input

The only line of the input contains integer n (0 ≤ n ≤ 1018) — the number of Ayrat's moves.

Output

Print two integers x and y — current coordinates of Ayrat coordinates.

Examples

Input

3


Output

-2 0


Input

7


Output

3 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void pr(vector<T> &v) {
  for (int i = 0; i < (int)(v).size(); i++) cout << v[i] << " ";
  cout << '\n';
  ;
}
template <typename T>
void pr(vector<vector<T>> &v) {
  for (int i = 0; i < (int)(v).size(); i++) {
    pr(v[i]);
  }
}
template <typename T>
void re(T &x) {
  cin >> x;
}
template <typename T>
void re(vector<T> &a) {
  for (int i = 0; i < (int)(a).size(); i++) re(a[i]);
}
template <class Arg, class... Args>
void re(Arg &first, Args &...rest) {
  re(first);
  re(rest...);
}
template <typename T>
void pr(T x) {
  cout << x << '\n';
  ;
}
template <class Arg, class... Args>
void pr(const Arg &first, const Args &...rest) {
  cout << first << " ";
  pr(rest...);
  cout << '\n';
  ;
}
void ps() {
  cout << '\n';
  ;
}
template <class T, class... Ts>
void ps(const T &t, const Ts &...ts) {
  cout << t;
  if (sizeof...(ts)) cout << " ";
  ps(ts...);
}
const long long MOD = 998244353;
long double PI = 4 * atan(1);
long double eps = 1e-12;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  long long lo = 0;
  long long hi = 1000000000;
  for (int i = 0; i < 100; i++) {
    long long mid = (lo + hi) / 2;
    if (3 * (mid) * (mid + 1) < n) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  long long back = 3 * (lo) * (lo + 1) - n;
  pair<int, int> diff[6] = {{-1, -2}, {-2, 0}, {-1, 2},
                            {1, 2},   {2, 0},  {1, -2}};
  pair<int, int> start = {lo * 2, 0};
  int cnt = 0;
  while (back > 0) {
    start.first += min(back, lo) * diff[cnt].first;
    start.second += min(back, lo) * diff[cnt].second;
    cnt++;
    back -= lo;
  }
  cout << start.first << " " << start.second << '\n';
  ;
}