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)

fun main(args: Array<String>) = yorukatsu9d()

fun yorukatsu9d() {
    val (n, x) = readLine()!!.split(' ').map { it.toLong() }

    val pattyAndBans = LongArray(51)
    pattyAndBans[0] = 1
    for (i in 0 until pattyAndBans.lastIndex) pattyAndBans[i + 1] = pattyAndBans[i] * 2 + 3

    val bans = LongArray(51)
    bans[0] = 1
    for (i in 0 until bans.lastIndex) bans[i + 1] = bans[i] * 2 + 1

    fun dfs(lv: Int, pos: Long): Long {
        if (lv == 0) return 1
        if (pos == 1L) return 0
        if (pos == pattyAndBans[lv]) return bans[lv]
        val mid = pattyAndBans[lv] / 2 + 1
        return when {
            pos < mid -> {
                dfs(lv - 1, pos - 1)
            }
            pos == mid -> {
                bans[lv - 1] + 1
            }
            else -> {
                dfs(lv - 1, pos - mid) + bans[lv - 1] + 1
            }
        }
    }

    val answer = dfs(n.toInt(), x)

    println(answer)
}