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,
        x: usize,
    }

    let mut burger = vec![0; n+1];
    let mut p = vec![0; n+1];

    p[0] = 1;
    burger[0] = 1;
    for i in 1..n+1 {
        p[i] = 2 * p[i - 1] + 1;
        burger[i] = 2 * burger[i - 1] + 3;
    }

    let ans = f(n, x, &burger, &p);

    println!("{}", ans);
}

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