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 fx[4] = {1, -1, 0, 0};
long long fy[4] = {0, 0, 1, -1};
int main() {
  long long n;
  while (cin >> n) {
    long long d = n % 7;
    n = n / 7;
    if (d == 0)
      cout << n * 2 << " " << n * 2 << "\n";
    else if (d == 1)
      cout << n * 2 << " " << (n + 1) * 2 - 1 << "\n";
    else if (d < 6)
      cout << n * 2 << " " << (n + 1) * 2 << "\n";
    else
      cout << n * 2 + 1 << " " << (n + 1) * 2 << "\n";
  }
  return 0;
}
