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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program atcoder
    implicit none
    integer(8) :: a(0:51),b(0:51),i,ans,n,x
    read*,n,x
    ans = 0
    a(0) = 1
    b(0) = 1
    do i=1,50
        a(i) = a(i-1) * 2 + 3
        b(i) = b(i-1) * 2 + 1 
    enddo
    call dfs(n,x)
    write(*,'(I0)')ans
!----------------------functions--------------------------
contains

recursive subroutine dfs(nn,xx)
    implicit none
    integer(8) :: nn,xx
    if(nn==0 .and. xx>=1)then
        ans = ans + 1
    else if(xx<=1)then
        ans = ans
    else if(xx>a(nn)/2)then
        ans = ans+1+b(nn-1)
        call dfs(n-1,xx-a(nn)/2-1)
    else if(xx==a(nn)/2)then
        ans = ans + b(n-1)
    else
        call dfs(nn-1,xx-1)
    end if
    return
end subroutine dfs

end program atcoder