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 <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using dbl = double;
using pii = pair<int, int>;
using pl4 = pair<ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;
using vd = vector<dbl>;
using vvd = vector<vd>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize
#define tp top()

#define p_q priority_queue

#define REP(i,a) for(int i=0;i<(a);i++)
#define REP1(i,a) for(int i=1;i<=(a);i++)
#define rREP(i,a) for(int i=(a-1);i>=0;i--)
#define rREP1(i,a) for(int i=(a);i>0;i--)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define ROR(v,i) for(auto &(i):(v))
#define IOTA(a,n) iota((a).be,(a).en,(n))
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define Imp(a) cout <<((a)?"Possible":"Impossible")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sak cout <<endl;
#define sas cout <<" ";
#define sat cout <<"\t";
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define c2l(a) ((ll)(a-48))
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2c(a) ((char)(a+48))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define Pow10(a) ((ll)(pow(10.0,double(a))))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define vin(v) ROR((v),(i)){cin >>(i);};
#define vllin(v,N) vll (v)((N));vin(v);
#define vsin(v,N) vs (v)((N));vin(v);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))
#define powll(a,b) (ll)(pow((double)(a),(double)(b)))
#define Triangle(x1,y1,x2,y2,x3,y3) (((x1)-(x2))*((y1)-(y3))-((x1)-(x3))*((y1)-(y2)))

#define int ll

const ll MOD = 1e9 + 7;
const ll N = 1e5 + 20;
//const ll MOD = 998244353;
//const ll MOD = 9007199254740881;
const ll INF = 1LL << 60;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); }
ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

ll inverse[N];

void init_inverse() {
	inverse[1] = 1;
	for (int i = 2; i < N; i++) inverse[i] = (MOD - MOD / i) * inverse[MOD%i] % MOD;
}

pl4 Bezout(ll a, ll b) {
	if (b != 0) {
		pl4 xy;
		xy = Bezout(b, a%b);
		return mp(xy.se, xy.fi - ((a / b)*xy.se));
	}
	if (b == 0) {
		return mp(1, 0);
	}
}
pl4 Bez(ll a, ll b, ll c) {
	pl4 xy;
	ll x, y, z, gc;
	xy = Bezout(a, b);
	gc = gcd(a, b);
	if (c%gc != 0) return mp(-1, -1);
	x = xy.fi*(c / gc); y = xy.se*(c / gc);
	if (x < 0) z = rup(-x, (b / gc));
	if (x >= 0) z = -x / (b / gc);
	x += z * (b / gc);
	y -= z * (a / gc);
	return mp(x, y);
}

void salv(vll v) {
	say("{");
	FOR(i, 0, v.sz - 1) {
		say(v[i]);
		if (i != v.sz - 1) say(",");
	}
	sal("}")
}

ll DigS10(ll n) {
	ll m = 0;
	FOR(i, 0, DigN10(n) - 1) {
		m += (ll)((llabs(n) % (ll)(pow(10.0, (double)(i + 1)))) / (ll)(pow(10.0, (double)i)));
	}
	return m;
}

ll isP(ll n) {
	if (n <= 1) return 0;
	FOR(i, 2, (ll)sqrt(n)) {
		if (n%i == 0) return 0;
	}
	return 1;
}

vll FactM(1, 1);
vll FactMI(1, 1);

ll PowM(ll a, ll b) {
	ll ans = 1, x = (a%MOD);
	FOR(i, 0, DigN2(b) - 1) {
		if (Dig2(b, i) == 1) ans = (ans*x) % MOD;
		if (i != (DigN2(b) - 1)) x = (x*x) % MOD;
	}
	return ans;
}

void CFactM(ll n) {
	if (FactM.sz <= n) {
		FOR(i, FactM.sz, n) {
			FactM.pb((FactM[i - 1] * (i%MOD)) % MOD);
		}
	}
	return;
}

void CFactMI(ll n) {
	CFactM(n);
	if (FactMI.sz < (n + 1)) FactMI.res(n + 1, -1);
	if (FactMI[n] == -1) FactMI[n] = PowM(FactM[n], MOD - 2);
	rFOR(i, 1, n - 1) {
		if (FactMI[i] != -1) break;
		FactMI[i] = ((FactMI[i + 1] * ((i + 1) % MOD)) % MOD);
	}
	return;
}

ll CombM(ll n, ll k) {
	if ((n < 0) || (k < 0)) return 0;
	if (n < k) return 0;
	if (n + 1 > FactMI.sz) CFactMI(n);
	return ((((FactMI[k] * FactMI[n - k]) % MOD)*FactM[n]) % MOD);
}

ll LIS(vll v, ll m) {
	if (v.sz > 0) {
		ll ans = 0;
		vll dp(v.sz, INF);
		FOR(i, 0, v.sz - 1) {
			dp[m ? UPB(dp, v[i]) : LOWB(dp, v[i])] = v[i];
		}
		FOR(i, 0, v.sz - 1) {
			if (dp[i] == INF) break;
			ans++;
		}
		return ans;
	}
	else {
		return 0;
	}
}

signed main() {
  ll x,r;
  cin >> x;
  r=x/11*2;
  x%=11;
  if(x==0){
    cout << r;
  } else if (x<=6) {
    cout << r+1;
  } else {
    cout << r+2;
  }
  return 0;
}