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;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    int s = n;
    if (n <= 5) {
      printf("0 %d\n", min(n, 2));
    } else {
      int sum = 0, sum1 = 2;
      while (s > 0) {
        if (s - 7 >= 0) {
          s -= 7;
          sum += 2;
        } else if (s == 6) {
          sum++;
          s -= 6;
        } else
          s -= 5;
      }
      n -= 2;
      while (n > 0) {
        if (n - 7 >= 0) {
          n -= 7;
          sum1 += 2;
        } else if (n == 6) {
          sum1++;
          n -= 6;
        } else
          n -= 5;
      }
      printf("%d %d\n", sum, sum1);
    }
  }
  return 0;
}
