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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
ll len[55],p[55];//分别存放Burger的长度和肉饼的个数  
ll slove(ll n,ll x)
{
	if(x<=n) return 0;
	else if(x>=len[n]-n) return p[n];
	else if(x==len[n]/2+1) return p[n]/2+1;
	else if(x<len[n]/2+1) return slove(n-1,x-1);
	else return p[n]/2+1+slove(n-1,x-len[n]/2-1);
}
int main()
{
	len[1]=5;
	p[1]=3;
	for(int i=2;i<=51;i++) 
	{
		len[i]=2*len[i-1]+3;
		p[i]=2*p[i-1]+1;
	}
	ll n,x;
	cin>>n>>x;
	cout<<slove(n,x)<<endl;
	return 0;
}
/*
BPPPB  1级   5  3

BBPPPBPBPPPBB 2级  2*5+3   13  7

BBBPPPBPBPPPBBPBBPPPBPBPPPBBB 3级    2*(2*5+3)+3   29  15

BBBBPPPBPBPPPBBPBBPPPBPBPPPBBBPBBBPPPBPBPPPBBPBBPPPBPBPPPBBBB 4级  2*29+3   31
*/