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>
int main() {
  int i, j, k, l, m;
  while (scanf("%d", &m) == 1) {
    if (m <= 2)
      printf("0 %d\n", m);
    else if (m > 2 && m <= 5)
      printf("0 2\n");
    else if (m == 6)
      printf("1 2\n");
    else if (m == 7)
      printf("2 2\n");
    else {
      i = m / 7;
      j = m % 7;
      if (j <= 2)
        k = (i * 2) + j;
      else if (j > 2 && j < 6)
        k = (i * 2) + 2;
      else if (j == 6)
        k = (i * 2) + 2;
      if (j < 6)
        printf("%d %d\n", i * 2, k);
      else if (j == 6)
        printf("%d %d\n", (i * 2) + 1, k);
    }
  }
  return 0;
}
