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>
#include <stdlib.h>
#define int long long

int all(int N){
	return ((int)1 << (N + 1)) - 1;
}

int solve(int N, int X){
	if(N == 0){
		if(X == 0){
			return 0;
		}
		else{
			return 1;
		}
	}
	else{
		if(X <= 1){
			return 0;
		}
		else if(X <= ((int)1 << (N + 1)) - 2){
			return solve(N - 1, X - 1);
		}
		else if(X == ((int)1 << (N + 1)) - 1){
			return all(N - 1) + 1;
		}
		else if(X <= ((int)1 << (N + 2)) - 4){
			return all(N - 1) + 1 + solve(N - 1, X - (((int)1 << (N + 1)) - 1));
		}
		else{
			return 2 * all(N - 1) + 1;
		}
	}
}

signed main(){
	int N, X, i;
	scanf("%lld%lld", &N, &X);
	printf("%lld\n", solve(N, X));
	return 0;
}