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)

use proconio::input;

fn main() {
    input! {
        n: usize,
        x: usize,
    };
    let mut l = vec![];
    l.push((1_usize, 1_usize));
    for i in 1..=n {
        l.push((l[i - 1].0 * 2 + 3, l[i - 1].1 * 2 + 1));
    }
    // for (i, l_i) in l.iter().enumerate() {
    //     println!("[{}]: {:?}", i, l_i);
    // }

    let mut s = 0;
    let mut o = 0;
    let mut c = n;
    loop {
        let m = l[c].0 / 2 + 1;
        if x == 1 + o {
            break;
        } else if x == m + o {
            s += l[c - 1].1 + 1;
            break;
        } else if x == l[c].0 + o {
            s += l[c].1;
            break;
        } else if x < m + o {
            c -= 1;
            o += 1;
        } else if x > m + o {
            s += l[c - 1].1 + 1;
            o += m;
            c -= 1;
        }
    }
    let ans = s;
    println!("{}", ans);
}