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>

int main()
{
    long long int n,x,i,j,b[51]={0},p[51]={0},ans=0,num[51]={0};
	int flag=0;
	
	scanf("%lld %lld",&n,&x);
	for(i=0;i<=n;i++)
	{
		if(i==0)p[i]=1;
		else
		{
			b[i]=b[i-1]*2+2;
			p[i]=p[i-1]*2+1;
		}
		num[i]=b[i]+p[i];
	}
	for(j=n;j>=0;j--)//にぶたん
	{
		if(x<=num[j]/2) x--;
		else if(x==(num[j]/2)+1)
		{
			ans+=(p[j-1]+1);
			break;
		}
		else
		{
			ans+=(p[j-1]+1);
			x-=(num[j-1]+2);
		}
	}
	
	
	printf("%lld",ans);
}