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::vec::Vec;

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 main() {
    let (n, x) = {
        let v: Vec<i64> = read_vec();
        (v[0], v[1])
    };

    let mut a = vec![1];
    let mut p = vec![1];

    for i in 0..n as usize {
        let t = a[i];
        let s = p[i];
        a.push(t * 2 + 3);
        p.push(s * 2 + 1);
    }

    let ans = f(&a, &p, n as usize, x);
    println!("{}", ans);
}

fn f(a: &Vec<i64>, p: &Vec<i64>, n: usize, x: i64) -> i64 {
   if n == 0 {
      return if x <= 0 { 0 } else { 1 };
   } else if x <= 1 + a[n-1] {
      return f(&a, &p, n-1, x-1);
   } else {
       return p[n-1] + 1 + f(&a, &p, n-1, x - 2 - a[n-1]);
   }
}