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: Wrong Answer)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;

#[fastout]
fn main() {
    input! {
    n: usize,
    x: usize
    }
    let ans = dfs(n, x);
    println!("{}", ans);
}

fn dfs(k: usize, x:usize) -> usize{
    if x <= k {
        return 0
    }
    let d = 2usize.pow(k as u32 + 2) - 3;
    if x == d {
        return 2usize.pow(k as u32) + 1
    }
    let mut res = 0;
    if d / 2 < x {
        res += 2usize.pow(k as u32);
        res += dfs(k - 1, x - d / 2 - 1);
    } else {
        res += dfs(k - 1, x - 1)
    }
    res
}