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;
const long long linf = 1000000000000000043;
const int inf = 1000000043;
void pause(bool a = true) {}
template <class T>
inline T prod(pair<T, T> a, pair<T, T> b, pair<T, T> c = make_pair(0, 0)) {
  return (a.first - c.first) * (b.second - c.second) -
         (a.second - c.second) * (b.first - c.first);
}
template <class T>
inline bool intriangle(pair<T, T> a, pair<T, T> b, pair<T, T> c, pair<T, T> d) {
  return (abs(prod(a, b, c)) ==
          abs(prod(a, b, d)) + abs(prod(a, c, d)) + abs(prod(b, c, d)));
}
template <class T>
inline int sg(T a) {
  return (a < 0 ? -1 : a ? 1 : 0);
}
long long rng(long long a, long long b) {
  return (((((long long)rand()) << 16) + rand()) % (b - a + 1)) + a;
}
bool inConvPoly(const vector<pair<long long, long long>> &a,
                pair<long long, long long> b) {
  long long l = 1;
  long long r = a.size() - 1;
  if (sg(prod(a[l], b, a[0])) * sg(prod(a[r], b, a[0])) >= 0) return false;
  if (intriangle(a[l], a[r], a[0], b)) return true;
  while (l != r - 1)
    (prod(a[(l + r) / 2], b, a[0]) < 0) ? l = (l + r) / 2 : r = (l + r) / 2;
  return sg(prod(b, a[r], a[l])) * sg(prod(b, a[0], a[l])) < 0;
}
template <class T>
T mex(vector<T> a) {
  set<T> b(a.begin(), a.end());
  for (int i = 0;; ++i)
    if (!b.count(i)) return i;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
template <class T>
inline const T sqr(const T &x) {
  return x * x;
}
unsigned long long powi(unsigned long long a, unsigned long long b,
                        long long mod = 1000000007) {
  if (b < 2) return b ? a : 1;
  return (powi((a * a) % mod, b / 2, mod) * (b % 2 ? a : 1)) % mod;
}
bool isn(char c) { return ('0' <= c && c <= '9'); }
template <class T>
bool read(T &out) {
  int c = getchar(), sign = 1;
  T num = 0;
  for (; !isn(c) && c != '-'; c = getchar())
    if (c == EOF) return false;
  if (c == '-') {
    sign = -1;
    c = getchar();
  }
  if (c == EOF) return false;
  for (; isn(c); c = getchar()) num = num * 10 + c - '0';
  out = sign * num;
  return true;
}
long long time_limit = 1000000;
long long start = clock();
bool time_break() {
  int g = clock();
  return (clock() - start > time_limit * 0.8);
}
long long fact(long long n, long long mod = linf) {
  long long g = 1;
  for (int i = 0; i < (n); i++) g *= (i + 1);
  return g;
}
void init(long long tl = 1000000) {
  ios::sync_with_stdio(false);
  srand(time(0));
  time_limit = tl;
}
template <class T>
void maxz(T &a, T b) {
  a = max(a, b);
}
template <class T>
void minz(T &a, T b) {
  a = min(a, b);
}
long long mod7 = 1000000007;
long long mod9 = 1000000009;
bool isprime(long long n) {
  if (n < 2) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
long long randLL() {
  return (((long long)rand()) << 47) + (((long long)rand()) << 31) +
         (((long long)rand()) << 15) + (((long long)rand()) >> 1);
}
bool isprimeL(long long n, int it = 50) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (int i = 0; i < (it); i++) {
    long long x = randLL() % (n - 1) + 1;
    if (gcd(x, n) != 1) return 0;
    if (powi(x, n - 1, n) != 1) return 0;
  }
  return 1;
}
bool Check(long long a, long long b) {
  bool d1 = a * 1.0 >= (b * 1.0) * (b * 1.0 + 1) * 3;
  bool d2 = a >= b * (b + 1) * 3;
  return !(a * 1.0 >= (b * 1.0) * (b * 1.0 + 1) * 3 && a >= b * (b + 1) * 3);
}
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long l = 0, r = 1e13;
  while (l != r - 1) {
    long long m = (l + r) / 2;
    if (Check(n, m))
      r = m;
    else
      l = m;
  }
  r = l;
  n -= r * (r + 1) * 3;
  long long x = 2 * r;
  r++;
  long long y = 0;
  y += 2 * min(n, 1LL);
  x += 1 * min(n, 1LL);
  n -= 1 * min(n, 1LL);
  y += 2 * min(n, r - 1);
  x -= min(n, r - 1);
  n -= min(n, r - 1);
  x -= 2 * min(n, r);
  n -= min(n, r);
  y -= 2 * min(n, r);
  x -= min(n, r);
  n -= min(n, r);
  y -= 2 * min(n, r);
  x += min(n, r);
  n -= min(n, r);
  x += 2 * min(n, r);
  n -= min(n, r);
  y += 2 * min(n, r);
  x += min(n, r);
  n -= min(n, r);
  cout << x << " " << y << endl;
  pause();
  return 0;
}