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: Accepted)

use std::cmp::min;
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
fn main() {
    let (n, x) = {
        let t = read_vec::<i64>();
        (t[0], t[1])
    };
    let ans = solve(n,x);
    println!("{}",ans);
}

fn solve(n:i64,x:i64) -> i64 {
    if n == 0 {
        return if x >= 1 { 1 } else { 0 };
    }

    let mut sum = 1 ;
    for _ in 0..n {
        sum = 3 + sum * 2;
    }
    if x == 1 {
        0
    } else if x <= sum /2 {
        solve(n-1,x-1)
    } else if x == sum /2 + 1 {
        solve(n-1,x-1) + 1
    } else {
        let mut p = 1;
        for _ in 0..n-1 {
            p = 1 + p *2;
        }
        solve(n-1,x - sum/2 - 1) + 1 + p
    }

}