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::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
  read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
  match n {
    0 => a,
    _ => {
      a.push(read::<T>());
      read_n_logic(n - 1, a)
    }
  }
}

use std::cmp::{max, min};
use std::collections::BinaryHeap;
use std::collections::HashMap;

fn main() {
  let mut n: usize = read();
  let mut x: usize = read();
  let mut ans = 0;
  // ;ans
  let mut p = vec![0; 51];
  let mut t = vec![0; 51];
  p[1] = 3;
  t[1] = 5;
  for i in 2..51 {
    p[i] = p[i - 1] * 2 + 1;
    t[i] = t[i - 1] * 2 + 3;
  }

  while 0 < n {
    let atumi = t[n - 1];
    let pathi = p[n - 1];

    if n == 1 {
      if x == 1 {
        ans += 1;
      } else if 1 < x && x <= 4 {
        ans += x - 1;
      } else {
        ans += 3
      }
      break;
    }

    if x == 1 {
      ans = 0;
      break;
    } else if x == atumi + 2 {
      ans = pathi + 1;
      break;
    } else if x == atumi * 2 + 3 {
      ans = pathi * 2 + 1;
      break;
    } else if 1 < x && x <= atumi + 1 {
      n -= 1;
      x -= 1;
    } else {
      n -= 1;
      x -= 1 + atumi;
      ans += pathi + 1;
    }
  }

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