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>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

long count_p(long X,int L, long* num, long* P, long tmp){
    long ans;
    if(L==0) ans=tmp+P[0];
    else{
        if(X==1) ans=tmp+0;
        else if(X==num[L]/2+1) ans=tmp+P[L-1]+1;
        else if(X==num[L]) ans=tmp+P[L];

        else if(X<(num[L]/2+1)){
            ans=count_p(X-1,L-1,num,P,tmp+0);
        }
        else{
            ans=count_p(X-num[L-1]-2,L-1,num,P,tmp+P[L-1]+1);
        }
    }

    return ans;
}

int main(void){
    int N;
    long X;
    scanf("%d%ld",&N,&X);

    //num[i]...レベルiバーガーが何枚か
    //P[i]...レベルiバーガーが何枚のパティがあるのか
    long num[N+1];
    long P[N+1];
    num[0]=P[0]=1;
    for(int i=1;i<=N;i++){
        num[i]=2*num[i-1]+3;
        P[i]=2*P[i-1]+1;
    }

    long ans=0;
    ans=count_p(X,N,num,P,ans);

    printf("%ld\n",ans);

    return 0;
}

//0 P
//1 B0P0B
//2 B1P1B = B B0P0B P B0P0B B
//3 B2P2B = B B B0P0B P B0P0B B P B B0P0B P B0P0B B B
//4 B B B B0P0B P B0P0B B P B B0P0B P B0P0B B B P B B B0P0B P B0P0B B P B B0P0B P B0P0B B B B