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>

#define LEN 51
unsigned long long a[LEN], p[LEN];

unsigned long long f(int N, unsigned long long X){
    if(X == 1){
        return 0;
    }
    else if(1 < X && X <= a[N-1] + 1){
        return f(N-1, X-1);
    }
    else if(a[N-1] + 1 < X && X == a[N-1] + 2){
        return p[N-1] + 1;
    }
    else if(a[N-1] + 2 < X && X <= 2 + 2*a[N-1]){
        return p[N-1] + 1 + f(N-1, X-2-a[N-1]);
    }
    else {
        return p[N];
    }
    
}

int main()
{
    int N, i;
    unsigned long long X;
    fscanf(stdin, "%d %lld", &N, &X);

    a[0] = 1; p[0] = 1;
    for(i = 1; i <= N; i++){
        a[i] = 2*a[i-1] + 3;
        p[i] = 2*p[i-1] + 1;
    }

    fprintf(stdout, "%lld\n", f(N, X));
}