On the planet Mars a year lasts exactly n days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 1 000 000) — the number of days in a year on Mars.

Output

Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.

Examples

Input

14


Output

4 4


Input

2


Output

0 2

Note

In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .

In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > 1000000007) x %= 1000000007;
    }
    y = (y * y);
    if (y > 1000000007) y %= 1000000007;
    b /= 2;
  }
  return x;
}
int main() {
  int n, i;
  int x, y;
  x = y = 0;
  cin >> n;
  x = min(2, n);
  for (i = 3; i <= n; i++) {
    if (i % 7 == 1 || i % 7 == 2) {
      x++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (i % 7 == 6 || i % 7 == 0) {
      y++;
    }
  }
  cout << y << " " << x << endl;
  return 0;
}
