/* FILE: FrameStewartConjecture.cpp   last change: 16-Sep-2013   author: Romeo Rizzi
 * a solver for problem FrameStewartConjecture based on dynamic programming
 */

//#define NDEBUG   // NDEBUG definita nella versione che consegno
#include <cassert>
#ifndef NDEBUG
#  include <iostream>  // uso di cin e cout non consentito in versione finale
#endif
#include <fstream>

using namespace std;

const int MAX_N = 10000; int n; 
const int MAX_K = 30; int k;
const int OVERFLOW = -1;
long int FS[MAX_N+1][MAX_K+1]; // FS[i][j] = the minimum number of moves in order to transport the tower according to a strategy obeying the Frame-Stewart conditions.


long int min (long int a, long int b) { return (a<b)? a : b; }

int main() {
  ifstream fin("input.txt"); assert( fin );
  fin >> n >> k;
  fin.close();

  // casi base:
  for(int j = 3; j <= k; j++) {
    FS[0][j] = 0; // nessun disco e' caso base 
    for(int i = 1; i < j; i++) // meno dischi di pioli e' caso base
      FS[i][j] = 1 +2*(i-1);
  }
  for(int i = 1; i <= n; i++)
    if( FS[i-1][3] == OVERFLOW )  FS[i][3] = OVERFLOW;
    else { 
      FS[i][3] = 1+2*FS[i-1][3]; // 3 pioli (classic Hanoi) e' caso base
      if ( FS[i][3] < FS[i-1][3] )  FS[i][3] = OVERFLOW; // overflow check
    }
  // step:
  for(int j = 4; j <= k; j++)
    for(int i = j; i <= n; i++) {
      FS[i][j] = OVERFLOW;
      for(int tip = 1; tip < i; tip++) {
        if( FS[tip][j] == OVERFLOW ) continue;
        if( FS[i-tip][j-1] == OVERFLOW ) continue;
        long int tmp = 2*FS[tip][j] + FS[i-tip][j-1];
        if( tmp < FS[i-1][j] ) continue;   // se overflow nella computazione di tmp
        if( FS[i][j] == OVERFLOW ) FS[i][j] = tmp;
        if( tmp < FS[i][j] ) FS[i][j] = tmp;
      }
    }

  ofstream fout("output.txt");
  fout << FS[n][k] << endl;
  fout.close();
  //few lines of code to experimentally determine the border:
  // for(int j = 3; j <= k; j++) {
  //   int i = 0;
  //   while( (FS[i+1][j] != OVERFLOW) && (i < n) ) i++;
  //   cout << "with " << j << " poles we get till n = " << i << endl; 
  // }

  return 0;
}