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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.*
import java.util.*

fun solve(N: Int, X: Long) {
    val sizes = LongArray(51)
    val patties = LongArray(51)
    sizes[0] = 1
    patties[0] = 1
    for (i in 1..N) {
        sizes[i] = 2 * sizes[i - 1] + 3
        patties[i] = 2 * patties[i - 1] + 1
    }
    fun pattyAmount(n: Int, x: Long): Long {
        if (x == 0L) return 0L
        if (n == 0) return 1L
        if (x == 1L) return 0L
        if (x >= sizes[n]) return patties[n]
        if (x >= (sizes[n] + 1) / 2) {
            return (patties[n] + 1) / 2 + pattyAmount(n, x - (sizes[n] + 1) / 2 + 1)
        }
        return pattyAmount(n - 1, x - 1)
    }
    println(pattyAmount(N, X))
    return
}

fun main(args: Array<String>) {
    fun StringArray(size: Int, init: (Int) -> String = { "\u0000" }): Array<String> {
        return Array<String>(size, init)
    }
    class Scanner(stream: InputStream) {
        private val reader = BufferedInputStream(stream).bufferedReader()
        private var st: StringTokenizer? = null
        fun next(): String {
            while (!(st?.hasMoreTokens() ?: false)) st = StringTokenizer(reader.readLine())
            return st!!.nextToken()
        }
    }
    val sc = Scanner(System.`in`)
    val N = sc.next().toInt()
    val X = sc.next().toLong()
    solve(N, X)
}