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: Wrong Answer)

use std::io::*;

fn rec(i: usize, x: u64, size: &Vec<u64>, pat: &Vec<u64>) -> u64 {
    if i == 0 {
        return 1;
    }
    if x == 0 {
        return 0;
    }
    if x < (size[i] + 1) / 2 {
        return rec(i - 1, x - 1, &size, &pat);
    }
    pat[i - 1] + 1 + rec(i - 1, x - (size[i] + 1) / 2, &size, &pat)
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let x: u64 = itr.next().unwrap().parse().unwrap();

    let mut size: Vec<u64> = vec![0; n + 1];
    let mut pat: Vec<u64> = vec![0; n + 1];
    size[0] = 1;
    pat[0] = 1;
    for i in 0..n {
        size[i + 1] = size[i] * 2 + 3;
        pat[i + 1] = pat[i] * 2 + 1;
    }
    println!("{}", rec(n, x, &size, &pat));
}