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: Accepted)

#include<cstdio>
long long a[51];
long long p[51];
long long f(int n , long long x)
{
    if(n == 0)
        if(x <= 0)
            return 0;
        else
            return 1;
    else if(x <= 1 + a[n - 1])
        return f(n - 1, x - 1);
    else
        return p[n - 1] + 1 + f(n - 1 , x - 2 - a[n - 1]);
}
int main()
{
    int n;
    long long x;
   // freopen("burgers.in" , "r" , stdin);
    //freopen("burgers.out" , "w" , stdout);
    scanf("%d%lld" , &n , &x);
    p[0] = 1;
    a[0] = 1;
    for(int i = 1; i <= n ; i ++)
    {
        a[i] = a[i - 1] * 2 + 3;
        p[i] = p[i - 1] * 2 + 1;
    }
    printf("%lld" , f(n , x));
    return 0;

}