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 dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, -1, 1};
void arquivo() {
  freopen("", "r", stdin);
  freopen("", "w", stdout);
}
int main() {
  int n;
  scanf("%d", &n);
  int c1, c2;
  c1 = c2 = 0;
  for (int i = 0; i < n; ++i) {
    if (i % 7 == 0 || i % 7 == 1) c1++;
    if (i % 7 == 5 || i % 7 == 6) c2++;
  }
  printf("%d %d\n", c2, c1);
  return 0;
}
