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 Rust (1.15.1) (judge verdict: Wrong Answer)

fn solve() {
    let mut buf = String::new();
    let handle = std::io::stdin();
    handle.read_line(&mut buf).unwrap();

    let info: Vec<usize> = buf.trim().split_whitespace()
        .map(|a| a.parse().unwrap()).collect();

    let n = info[0];
    let x = info[1];
    let ans = f(n as isize, x as isize);
    println!("{}", ans);
}

fn f(n: isize, x: isize) -> isize {
    if n == 0 { return 1;}
    if n >= x { return 0;}
    if (burger_len(n) + 1) / 2 > x {
        f(n-1, x-1)
    } else {
        2_isize.pow((n+1) as u32) - 1
        - f(n-1, burger_len(n) - x - 1)
    }
}

fn burger_len(n: isize) -> isize {
    2_isize.pow((n+2) as u32) - 3
}
fn main() {
    solve();
}