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: Runtime Error)

#!/mnt/c/Users/moiki/bash/env/bin/python
N, X = map(int, input().split())

def calcl(n):
    l = 2 ** (n+2) - 3 
    return l
def calcp(n):
    p = 2 ** (n+1) - 1
    return p
    
L = [1]
P = [1]

for i in range(1, N+1):
    L.append( L[i-1] * 2 + 3)
    P.append( P[i-1] * 2 + 1)

def judge(n, x):
    if n == 0:
        if x == 1:
            return 1
        else:
            raise "error"
    # if x == 1:
    #     if n == 0:
    #         return "3"
    #     else:
    #         return "2"
    else:
        l = L[n]       # layers
        top = l-1
        bottom = 0
        mid = (l-1)//2 # index of middle

        if x == mid:
            return "3"
        elif top > x > mid:
            return "1"
        elif 0 < x < mid:
            return "0"            
        elif x == top:
            return "4"
        elif x == bottom:
            return "2"
        else:
            raise "No such pattern!"
        
    
def calc_pan(n,x, sum):
    c = judge(n,x)
    # print("calc_panK: n:{}, x:{}, c: {}".format(n,x,c))
    if  c == "2":
        return sum
    elif c == "4":
        return sum + 1 + P[n-1]
    elif c == "3":
        return sum + 1 
    elif c == "0":
        return sum + calc_pan(n-1, x-1, 0)
    elif c == "1":
        return sum + P[n-1] + 1 + calc_pan(n-1, x-L[n-1]-2, 0)





sum = calc_pan(N, X, 0)
print(sum)