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 proconio::input;

fn main() {
    input! {
        n: usize,
        m: i64,
    }

    let mut burger_size = vec![0i64; 51];
    let mut patty_count = vec![0i64; 51];
    burger_size[0] = 1;
    patty_count[0] = 1;

    for i in 1..=50 {
        burger_size[i] = burger_size[i - 1] * 2 + 3;
        patty_count[i] = patty_count[i - 1] * 2 + 1;
    }

    struct Ctx {
        burger_size: Vec<i64>,
        patty_count: Vec<i64>,
    }

    fn eat(e: &Ctx, n: usize, rem: i64) -> i64 {
        if n == 0 {
            1
        } else {
            let Ctx { burger_size, patty_count } = e;
            if rem <= 1 {
                0
            } else if rem <= 1 + burger_size[n - 1] {
                eat(e, n - 1, rem - 1)
            } else if rem == 1 + burger_size[n - 1] + 1 {
                patty_count[n - 1] + 1
            } else if rem <= 1 + burger_size[n - 1] + 1 + burger_size[n - 1] {
                patty_count[n - 1] + 1 + eat(e, n - 1, rem - 1 - burger_size[n - 1] - 1)
            } else {
                patty_count[n]
            }
        }
    }

    println!("{}", eat(&Ctx{ burger_size, patty_count }, n, m));
}