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: Runtime Error)

nx = gets.chomp!.split(" ")
nx = nx.map{|item| item=item.to_i}

hamburgers = Array.new(nx[0],[])
hamburgers[0] = [false]#bun is true, patty is false
for i in 1..nx[0]
	hamburgers[i-2] = nil
	if 1==nx[1] then
		hamburgers[i] = [true]
	elsif hamburgers[i-1].size+1>=nx[1] then
		hamburgers[i] = [true] + hamburgers[i-1] 
	elsif hamburgers[i-1].size+2==nx[1] then
		hamburgers[i] = [true] + hamburgers[i-1] + [false]
	elsif hamburgers[i-1].size*2+2>=nx[1] then
		hamburgers[i] = [true] + hamburgers[i-1] + [false] + hamburgers[i-1]
	else
		hamburgers[i] = [true] + hamburgers[i-1] + [false] + hamburgers[i-1] + [true]
	end
end

answer = hamburgers[nx[0]].take(nx[1]).count(false)

#puts hamburgers.inspect
puts answer