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>
#include <cmath>
#include <iostream>

using namespace std;

typedef unsigned long long ll;

ll p[55], nlen[55];

ll solve(ll n, ll x)
{
    if(x <= n) return 0;
    if(x >= nlen[n] - 1) return p[n];
    if(x == nlen[n] / 2 + 1) return (p[n] / 2 + 1);
    else if(x < nlen[n] / 2 + 1)
        return solve(n - 1, x - 1);
    else
        return p[n] / 2 + 1 + solve(n - 1, x - nlen[n] / 2 - 1);
}

int main()
{
    for(int i = 0; i <= 50; ++i)
    {
        p[i] = pow(2, i + 1) - 1;
        nlen[i] = pow(2, i - 1) * 8 - 3;
    }
    ll n, x;
    cin>>n>>x;
    cout<<solve(n, x)<<endl;
    return 0;
}