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)

allEatCache = {}
lenBurgerCache = {}


def nOfPatty(level, eat):
    if level == 0:
        return 1
    else:
        if level in allEatCache and eat >= lenBurger(level):
            return allEatCache[level]
        else:
            lenOfLowerLevel = lenBurger(level - 1)
            if eat <= 1:
                return 0
            if 2 <= eat <= 1 + lenOfLowerLevel:
                return nOfPatty(level - 1, eat - 1)
            if eat == 1 + lenOfLowerLevel + 1:
                return nOfPatty(level - 1, eat - 1) + 1
            else:
                tmp = nOfPatty(level - 1, eat - 1) + 1
                tmp += nOfPatty(level - 1, eat - 1 - lenOfLowerLevel - 1)
                allEatCache[level] = tmp
                return tmp


def lenBurger(level):
    if level == 0:
        return 1
    else:
        if level in lenBurgerCache:
            return lenBurgerCache[level]
        else:
            return 3 + lenBurger(level - 1) * 2


str = input().split()
n = int(str[0])
x = int(str[1])

print(nOfPatty(n, x))