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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Data.Int

main = do
    [n,x] <- map read . words <$> getLine :: IO [Int] 
    print $ eat x $ mkBurgerNum n

-- (total，pate)
mkBurgerNum :: Int -> [[Int]]
mkBurgerNum n 
    | n == 0 = [[1,1]]
    | otherwise = [2*m+3, 2*l+1]:x
    where 
        x = mkBurgerNum (n-1)
        m = x !! 0 !! 0
        l = x !! 0 !! 1

-- x mkBurgerNum 
eat :: Int -> [[Int]] -> Int 
eat x (berger:[]) = 1
eat x (berger:child_berger:xs)
    | x == 1 = 0
    | x > child_berger_num + 2 = child_pate_num + 1 + eat (x - 2 - child_berger_num) (child_berger:xs)
    | x == child_berger_num + 2 = 1 + child_pate_num
    | otherwise = eat (x-1) $ child_berger:xs
    where 
        child_berger_num = child_berger !! 0 
        child_pate_num = child_berger !! 1