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 Python (3.4.3) (judge verdict: Wrong Answer)

N, X = map(int, input().split())

# レベルLバーガーの厚さ, レベルLバーガーに含まれるパティの数
thick, patti = [1], [1]
for i in range(N):
    thick.append(2 * thick[-1] + 3)
    patti.append(2 * patti[-1] + 1)


def solve(n, x):
    if x == 1:  # 一番下のバン
        return 0

    # 一番下のバン + レベルL-1バーガーの途中まで
    elif 1 < x < 1 + thick[n - 1]:
        return solve(n - 1, x - 1)

    # 一番下のバン + レベルL-1バーガー + 真ん中のパティ
    elif x == 2 + thick[n - 1]:
        return patti[n - 1] + 1

    # 一番下のバン + レベルL-1バーガー + 真ん中のパティ + レベルL-1バーガーの途中まで
    elif 2 + thick[n - 1] < x < 2 * thick[n - 1] + 3:
        return patti[n - 1] + 1 + solve(n - 1, x - 1 - thick[n - 1] - 1)

    # 一番下のバン + レベルL-1バーガー + 真ん中のパティ + レベルL-1バーガー + 最後のバン
    else:
        return 2 * patti[n - 1] + 1


print(solve(N, X))