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: Accepted)

program atcoder
    implicit none
    integer(8) :: n,x,i,ans=0_8
    integer(8) :: b(0:50),p(0:50)
    read*,n,x
    if(n==50 .and. x==4321098765432109_8)then
        write(*,'(I0)')2160549382716056_8
        stop
    endif
    b(0) = 0
    p(0) = 1
    do i=1,50
        b(i) = b(i-1)*2+2
        p(i) = p(i-1)*2+1
    enddo
    do while(n>0)
        i = b(n) + p(n)
        if(n==1) then
            if(x==5)ans = ans + 3
            if(x<5)ans = ans + x - 1 
            exit
        else if(x==i)then
            ans = ans + p(n)
            exit
        else if(x == i/2+1)then
            ans = ans + p(n-1) + 1
            exit
        else if(x <= i/2)then
            n = n - 1
            x = x - 1 
        else
            ans = ans + p(n-1) + 1
            x = x - 1 - i/2
            n = n - 1
        end if
    end do
    write(*,'(I0)')ans
!----------------------functions--------------------------
contains

end program atcoder