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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
long long int vp[60],len[60],sum=0;


int main()
{
    int n;long long int x;
    scanf("%d %lld",&n,&x);
    vp[1]=3;len[1]=5;
    for(int i=2;i<=n;i++)
    {
        vp[i]=vp[i-1]*2+1;
    }
    for(int i=2;i<=n;i++)
    {
        len[i]=len[i-1]*2+3;
    }
    while(x>0)
    {
        if(n==0)
            break;
        if(x<=5)
        {
            if(x==5)
                x--;
            if(x>=2)
                sum+=x-1;

            break;
        }
        if(x>=(len[n]-1)/2+1)
            {x-=(len[n]-1)/2+1;

               sum+=vp[n-1]+1;
               n--;
            }
        else
        {
            x--;
            n--;
        }

    }
    printf("%lld",sum);

}