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())
X -= 1
p = [1]*(N+1)
b = [1]*(N+1)

def where_X(s, b, X, n):
  X -= s
  l = b[n]+1
  if X == 0:
    return 0
  elif 0 < X < l:
    return 1
  elif X == l:
    return 2
  elif l < X < 2*l:
    return 3
  elif X == 2*l:
    return 4


for i in range(1, N+1):
  p[i] = 2*p[i-1] + 1
  b[i] = 2*b[i-1] + 3

now_N = N-1
result = 0
start = 0
while True:
  lo_X = where_X(start, b, X, now_N)
  if lo_X == 0:
    break
  elif lo_X == 1:
    start += 1
  elif lo_X == 2:
    result += 1+p[now_N]
    break
  elif lo_X == 3:
    start += 2 + b[now_N]
    result += 1 + p[now_N]
  elif lo_X == 4:
    result += 1 + 2*p[now_N]
    break
  now_N -= 1

print(result)