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) = list(map(int, input().split()))

num = 1
pate = 1
for i in range(n):
    num = num * 2 + 3
    pate = pate * 2 + 1

def culc(l, idx, num, pate):
    npate = (pate - 1) // 2
    nnum = (num - 3) // 2
    mid = (num + 1) // 2
    if l == 1:
        if idx == 1:
            return 0
        elif idx == 2:
            return 1
        elif idx == 3:
            return 2
        elif idx == 4:
            return 3
        else:
            return 3
    elif mid == idx:
        return npate + 1
    elif mid < idx:
        return npate + 1 + culc(l - 1, idx - mid, nnum, npate)
    else:
        return culc(l - 1, idx - 1, nnum, npate)
print(culc(n, x, num, pate))