#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>	// require sort next_permutation count __gcd reverse etc.
#include <cstdlib>	// require abs exit atof atoi 
#include <cstdio>		// require scanf printf
#include <functional>
#include <numeric>	// require accumulate
#include <cmath>		// require fabs
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip>	// require setw
#include <sstream>	// require stringstream 
#include <cstring>	// require memset
#include <cctype>		// require tolower, toupper
#include <fstream>	// require freopen
#include <ctime>		// require srand
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

/*
	No.65 回数の期待値の練習

	No.72 の確率固定版。

	E[1]	= 1
	E[2]	= E[1]*P1 + 1 = E[1]/6 + 1
	E[3]	= E[2]*P1 + E[1]*P2 + 1 = (E[2] + E[1] )/6 + 1
	E[4]	= E[3]*P1 + E[2]*P2 + E[1]*P3 + 1 = (E[3] + E[2] + E[1] )/6 + 1
	E[5]	= E[4]*P1 + E[3]*P2 + E[2]*P3 + E[1]*P4 + 1 = (E[4] + E[3] + E[2] + E[1] )/6 + 1
	E[6]	= E[5]*P1 + E[4]*P2 + E[3]*P3 + E[2]*P4 + E[1]*P5 + 1 = (E[5] + E[4] + E[3] + E[2] + E[1] )/6 + 1
	E[7]	= E[6]*P1 + E[5]*P2 + E[4]*P3 + E[3]*P4 + E[2]*P5 + E[1]*P6 + 1
		= (E[6] + E[5] + E[4] + E[3] + E[2] + E[1] )/6 + 1

	問題文は x >= 7 のとき
	E[x] = E[x-1]*1/6 + E[x-2]*1/6 + E[x-3]*1/6 + E[x-4]*1/6 + E[x-5]*1/6 + E[x-6]*1/6 + 1 の誤植かな？

*/

const int MAX_K = 22;

double E[MAX_K];

int main()
{
	memset (E, 0., sizeof (E ) );
	ios_base::sync_with_stdio(0);
	E[1] = 1.;
	E[2] = E[1]/6. + 1.;
	E[3] = (E[2] + E[1] )/6. + 1.;
	E[4] = (E[3] + E[2] + E[1] )/6. + 1.;
	E[5] = (E[4] + E[3] + E[2] + E[1] )/6. + 1.;
	E[6] = (E[5] + E[4] + E[3] + E[2] + E[1] )/6. + 1.;
	for (int i = 7; i < MAX_K; i++ ){
		for (int j = 1; j <= 6; j++ ){
			E[i] += E[i-j]/6.;
		} // end for
		E[i] += 1.;
	} // end for

	int K; cin >> K;
	printf ("%.6lf\n", E[K] );
 
	return 0;
}