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 (1.42.0) (judge verdict: Accepted)

use proconio::input;


struct Burger {
    a: Vec<usize>,
    p: Vec<usize>,
}

impl Burger {
    fn new(n: usize) -> Self {
        let mut a = vec![1];
        let mut p = vec![1];
        for i in 0..n {
            a.push(2 * a[i] + 3);
            p.push(2 * p[i] + 1);
        }
        Self { a, p }
    }

    fn solve(&self, n: usize, x: isize) -> usize {
        if n == 0 {
            if x <= 0 {
                0
            } else {
                1
            }
        } else if x <= self.a[n-1] as isize + 1 {
            self.solve(n-1, x-1)
        } else {
            self.solve(n-1, x - self.a[n-1] as isize - 2) + self.p[n-1] + 1
        }
    }
}

fn main() {
    input! {
        n: usize,
        x: usize,
    }

    let burger = Burger::new(n);

    let ans = burger.solve(n, x as isize);
    
    println!("{}", ans);

}