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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
// D
fscanf(STDIN, '%d %d', $n, $x);
$cnt = [1];
$cnt_p = [1];
for ($i = 1; $i <= $n; $i++) {
    $cnt[] = $cnt[$i - 1] * 2 + 3;
    $cnt_p[] = $cnt_p[$i - 1] * 2 + 1;
}
echo f($n, $x);

function f($n, $x)
{
    global $cnt, $cnt_p;

    if ($n === 0) return min(1, $x);

    if ($x <= 1) return 0;
    if ($x <= ($cnt[$n] - 3) / 2 + 1) {
        return f($n - 1, $x - 1);
    }
    if ($x === ($cnt[$n] - 3) / 2 + 2) {
        return f($n - 1, $x - 2) + 1;
    }
    if ($x < $cnt[$n] - 1) {
        return $cnt_p[$n - 1] + 1 + f($n - 1, $x - (($cnt[$n] - 3) / 2 + 2));
    }
    return $cnt_p[$n - 1] * 2 + 1;
}