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 std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let x: usize = iterator.next().unwrap().parse().unwrap();

    let mut a: Vec<usize> = vec![1; n + 1];
    let mut p: Vec<usize> = vec![1; n + 1];
    for i in 1..=n {
        a[i] = 2 * a[i - 1] + 3;
        p[i] = 2 * p[i - 1] + 1;
    }

    f(n, x, &a, &p).to_string()
}

fn f(n: usize, x: usize , a: &Vec<usize>, p: &Vec<usize>) -> usize {
    if n == 0 {
        return 1;
    }

    if x == 1 {
        return 0;
    }

    if x < a[n - 1] + 2 {
        return f(n - 1, x - 1, a, p);
    }

    if x == a[n - 1] + 2 {
        return p[n - 1] + 1;
    }

    if x < a[n] {
        return p[n - 1] + 1 + f(n - 1, x - (a[n - 1] + 2), a, p);
    }

    return p[n]
}