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>) {
    val (N, X) = readLine()!!.split(" ").map{it.toLong()}

    val t = Array(60) { 0L }
    val p = Array(60) { 0L }

    t[0] = 1
    p[0] = 1
    for (i in 1 until t.size) {
        t[i] = t[i - 1] * 2 + 3
        p[i] = p[i - 1] * 2 + 1
    }

    fun rec(n: Int, x: Long): Long {
        if (t[n] <= x) {
            return p[n]
        }

        if (n == 0) {
            return 0L
        }

        var r = 0L

        if (x < t[n - 1] + 1) {
            r += rec(n - 1, x - 1)
        }

        if (x >= t[n - 1] + 1) {
            r += p[n - 1]
        }

        if (x >= t[n - 1] + 2) {
            r += 1L
        }

        if (x > t[n - 1] + 2) {
            r += rec(n - 1, x - t[n - 1] - 2)
        }

        return r
    }

    println(rec(N.toInt(), X))
}