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 <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12


int n;
llong x;
llong int b[60];
llong int p[60];

llong run(int n,llong x){
	if(n==0){
		return 1;
	}else{
		if(x<=1){
			return 0;
		}else if(x<=b[n-1]+1){
			return run(n-1,x-1);
		}else if(x==b[n-1]+2){
			return run(n-1,x-1)+1;
		}else if(x<=2*b[n-1]+2){
			return run(n-1,x-b[n-1]-2)+p[n-1]+1;
		}else if(x==2+b[n-1]+3){
			return p[n];
		}
	}
}


int main(void)
{
	//変数の宣言
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	llong int ans=0;
	int count=0;
	int temp,temp1,temp2;
	llong int max,min;
	int len;
	int sum=0;
	//データの読み込み
	scanf("%d %lld",&n,&x);
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	
	for(i=0;i<=50;i++){
		if(i==0){
			b[0]=1;
			p[0]=1;
		}else{
			b[i]=b[i-1]*2+3;
			p[i]=p[i-1]*2+1;
		}
	}
	
	ans=run(n,x);
	
//	printf("計算部分終了\n");
	//出力
	
	printf("%lld",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}