/*

     题意：汉诺塔问题变形，多了第四个盘子可以放前k个塔，然后n-k个是经典的汉诺塔问题，问最少操作次数

     递推+高精度+找规律：f[k]表示前k放在第四个盘子，g[n-k]表示经典三个盘子，2 ^ (n - k) - 1

             所以f[n] = min (f[k] * 2 + g[n-k])，n<=10000，所要要用高精度，另外打表能看出规律

 */

 /************************************************

 * Author        :Running_Time

 * Created Time  :2015-8-18 9:14:21

 * File Name     :UVA_10254.cpp

  ************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN =  + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 struct bign {

     short s[MAXN] , len ;

     bign () { memset ( s , , sizeof ( s ) ) ; len =  ; }

     bign operator = (const char *num) {

         len = strlen ( num ) ;

         for ( int i =  ; i < len ; i ++ ) s[i] = num[len-i-] - '' ;

         return *this ;

     }

     bign operator = (int num) {

         char s[MAXN];

         sprintf (s , "%d" , num);

         *this = s ;

         return *this ;

     }

     bign(const char *num) { *this = num ; }

     bign(int num) { *this = num ; }

     string str () const {

         string res ;

         res = "" ;

         for (int i = ; i < len; i ++) res = (char) (s[i] + '') + res ;

         if (res == "") res = '';

         return res ;

     }

     bign operator + (const bign& b) const {

         bign c ;

         c.len =  ;

         for(int i = , g = ; g || i < max (len, b.len); i ++) {

             int x = g ;

             if (i < len) x += s[i] ;

             if (i < b.len) x += b.s[i] ;

             c.s[c.len++] = x %  ;

             g = x /  ;

         }

         return c ;

     }

     void print() {

         for(int i = len - ; i >= ; i --) printf("%hd", s[i]);

         printf("\n");

     }

 }f[];

 int main(void)  {       //UVA 10254 The Priest Mathematician

     bign g = ; f[] = ;

     for (int i=, j=; i<=; j++, g=g+g)    {

         for (int k=; k<=j && i<=; k++,i++)    {

             f[i] = f[i-] + g;

         }

     }

     int n;

     while (scanf ("%d", &n) == )   {

         f[n].print ();

     }

     return ;

 }