Problem:
Score : 300 points

Problem Statement

Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7.

Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation:

Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward.

For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure.
If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward.
Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back.

Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.

Constraints

1 ≦ x ≦ 10^{15}

x is an integer.

Input

The input is given from Standard Input in the following format:

x

Output

Print the answer.

Sample Input 1

7

Sample Output 1

2

Sample Input 2

149696127901

Sample Output 2

27217477801

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cmath>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<iomanip>
#include<queue>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> ipair;
bool lessPair(const ipair& l, const ipair& r){return l.second < r.second;}
bool morePair(const ipair& l, const ipair& r){return l.second > r.second;}

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const ll MOD = 1e9 + 7;
// const long long INF = 1LL<<60;
void add(long long &a, long long b) { a += b; if (a >= MOD) a -= MOD; }
void sub(long long &a, long long b) { a -= b; if (a < 0) a += MOD; }
void mul(long long &a, long long b) { a *= b; a %= MOD; }
ll llmin(ll a, ll b) { if (a < b) return a; else return b; }
ll llmax(ll a, ll b) { if (a < b) return b; else return a; }
ll llabs(ll a) { if (a >= 0) return a; else return - a; }
ll llmodpow(ll a, ll n) {
   if (n == 0) return 1;
   ll tmp = llmodpow(a, n / 2);
   mul(tmp, tmp);
   if (n & 1) mul(tmp, a);
   return tmp;
}

int main() {
   ll x;
   cin >> x;
   ll ans = x / 11;
   ans *= 2;
   ll tmp = x % 11;
   if (tmp != 0) {
      if (tmp <= 6) ans += 1;
      else ans += 2;
   }
   cout << ans << endl;
   return 0;
}