Problem:
Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

A level-0 burger is a patty.

A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

1 \leq N \leq 50

1 \leq X \leq ( the total number of layers in a level-N burger )

N and X are integers.

Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

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

#include <stdio.h>

long pow(long a,int n){
  long answer = 1;
  for(int i = 0; i < n; ++i) answer *= a;
  return answer;
}

long eat(int n,long x){//n層下からxまでのpatty
  if(n == 1){
    if(x == 1) return 0;
    else if(x == 5) return 3;
    else return x - 1;
  }else if(n > 1){
    long p = pow(2,n+1) - 1; //真ん中のpの位置 = nバp数
    long pp = (p - 1) / 2; //n-1Burger
    long a = p - 2;        //n-1バーガーの層数
    long all = 2 * p - 1; //nBurgerの層数
    if(x == 1) return 0;  //bのみ
    else if(x < p) return eat(n-1,x-1); //下のburger
    else if(x == p) return 1 + eat(n-1,x-2);
    else if(x < all) return 1 + pp + eat(n-1,x - p);
    else if(x == all) return 1 + pp + eat(n-1,x - p - 1);
  }
}

int main(){
  int n;
  long x;
  scanf("%d %ld",&n,&x);
  long p = pow(2,n+1) - 1;//真ん中のpの位置
  long a = pow(2,n+1) - 3;//n-1バーガーの層数
  long answer = eat(n,x);
  printf("%ld\n",answer);
  return 0;
}