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>

long long pate(long long burger[], long long p[], int n, long long x);

int main(void)
{
	int n;
	long long x, p[51], burger[51], ans;
	scanf("%d%lld", &n, &x);
	p[0]=1;
	burger[0]=1;
	for (int i = 1; i <= n; ++i) p[i]=2*p[i-1]+1;
	for (int i = 1; i <= n; ++i) burger[i]=2*burger[i-1]+3;
	ans=pate(burger, p, n, x);
	printf("%d\n", ans);
	return 0;
}

long long pate(long long burger[], long long p[], int n, long long x)
{
	long long ans;
	if (x>=burger[n]-n)
	{
		ans=p[n];
	}
	else if (x>burger[n-1]+2)
	{
		x=x-burger[n-1]-2;
		ans=p[n-1]+pate(burger, p, n-1, x)+1;
	}
	else if (x==burger[n-1]+2)
	{
		ans=p[n-1]+1;
	}
	else if (x>n)
	{
		ans=pate(burger, p, n-1, x-1);
	}
	else
	{
		ans=0;
	}
	return ans;
}