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: Accepted)

import os, sys, re, math

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

memo_length = [0 for i in range(51)]
memo_patties = [0 for i in range(51)]
memo_length[0] = 1
memo_patties[0] = 1

for i in range(1,51):
    memo_length[i] = memo_length[i-1] * 2 + 3
    memo_patties[i] = memo_patties[i-1] * 2 + 1

def doit(level,x):
    if x == 0:
        return 0
    if x >= memo_length[level]:
        return memo_patties[level]

    tmp = memo_length[level-1]
    ret = 0

    if x >= 1:
        ret += doit(level-1, x-1)
    if x >= tmp + 2:
        ret += 1
        ret += doit(level-1, x-tmp-2)

    return ret

print(doit(N,X))