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 Ruby (2.3.3) (judge verdict: Wrong Answer)

N,X=gets.chomp.split.map(&:to_i)
b=[]
p=[]
p[0]=1
b[0]=0

1.upto(N) do |i|
    p[i] = 1+2*p[i-1]
    b[i] = 2+2*b[i-1]
end

def get_pb(n,m,s,p,b)
    if m == 1 
        s
    elsif m == 2 + p[n-1] + b[n-1]
        s + 1 + p[n-1]
    elsif m == 3 + 2 * (p[n-1] + b[n-1])
        s + 1 + 2* p[n-1]
    elsif m < 2 + p[n-1] + b[n-1]
        get_pb(n-1, m-1, s, p,b)
    else
        get_pb(n-1, m-2-p[n-1]-b[n-1], s+1+p[n-1], p,b)        
    end
end

puts get_pb(N,X,0,p,b)