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 <iostream>
#include <vector>
#include <algorithm>

const int MAX_N = 50;

long long thickness[MAX_N + 1];

void thickness_init()
{
    thickness[0] = 1;
    for (int i = 1; i < MAX_N + 1; i++) {
        thickness[i] = thickness[i-1] * 2 + 3;
    }
}

long long rec(int level, long long x) {
    if (x == 0) return 0;
    if (level == 0 && x == 1) return 1;

    long long ret = 0;

    if (1 <= x && x <= thickness[level-1]+1)
    {
        ret = rec(level-1, x-1);
    }
    else if (thickness[level-1]+1 < x && x < thickness[level]) {
        ret = rec(level-1, thickness[level-1]) + 1 + rec(level-1, x - 2 - thickness[level-1]);
    }
    else if (x == thickness[level])
    {
        ret = 2 * rec(level-1, thickness[level-1]) + 1;
    }
    return ret;
}

int main()
{
    thickness_init();
    long long n, x;
    std::cin >> n >> x;
    std::cout << rec(n, x) << std::endl;
    return 0;
}