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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=55;
ll level,n,h[N],p[N];
inline ll solve(ll x,ll y)
{
	if(!x)return 1;
	if(y==1)return 0;
	if(y>1&&y<h[x-1])//在第一层
		return solve(x-1,y-1);//去掉第二层加上中间的
	if(y==h[x-1]+2)return p[x-1]+1;//在中间
	if(y>h[x-1]+2&&y<3+2*h[x-1])//在第二层
		return p[x-1]+1+solve(x-1,y-2-h[x-1]);//去掉第一层+中间的+下面的
	return p[x-1]*2+1;//最后正好
}
int main()
{
	cin>>level;
	cin>>n;
	h[0]=1;
	p[0]=1;
	for(int i=1;i<=50;i++)
	{
		h[i]=3+h[i-1]*2;
		p[i]=1+p[i-1]*2;
	}
	cout<<solve(level,n)<<endl;
	return 0;
}