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) = readInputLine().split(" ").map{it.toLong()}
    
    val a = LongArray(N.toInt() + 1)
    val p = LongArray(N.toInt() + 1)
    
    a[0] = 1L
    p[0] = 1L
    
    for (i in 1..N.toInt()) {
        a[i] = a[i - 1] * 2L + 3L
        p[i] = p[i - 1] * 2L + 1L
    }

    println(dfs(a, p, N.toInt(), X))
}

fun readInputLine(): String {
    return readLine()!!
}

fun dfs(a: LongArray, p: LongArray, N: Int, X: Long): Long {
    if (N == 0) {
        return X
    }
    return when {
        X == 1L -> 0L
        1L < X && X <= a[N - 1] + 1L -> dfs(a, p, N - 1, X - 1L)
        X == a[N - 1] + 2L -> p[N - 1] + 1L
        a[N - 1] + 2L < X && X <= a[N - 1] * 2L + 2L -> p[N - 1] + 1L + dfs(a, p, N - 1, X - a[N - 1] - 2L)
        else -> p[N - 1] * 2L + 1L
    }
}