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 n, m, k, kq, d;
long long a[1000005], b[1000005], c[1000005], kt[1000005], power[1000005];
string s1, s2, skq;
int main() {
  cin >> n;
  long k1 = 0, k2 = 0;
  for (int i = 3; i <= n; i++) {
    if (k1 < 5) {
      k1++;
      a[i]++;
    } else
      k2++;
    if (k1 == 5 && k2 == 2) {
      k1 = 0;
      k2 = 0;
    }
  }
  k1 = 0;
  k2 = 0;
  for (int i = 1; i <= n; i++) {
    if (k1 < 5) {
      k1++;
      b[i]++;
    } else
      k2++;
    if (k1 == 5 && k2 == 2) {
      k1 = 0;
      k2 = 0;
    }
  }
  long kq1 = 0, kq2 = 0;
  for (int i = 1; i <= n; i++)
    if (b[i] == 0) kq1++;
  for (int i = 1; i <= n; i++)
    if (a[i] == 0) kq2++;
  cout << kq1 << " " << kq2;
}
