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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php

fscanf(STDIN, "%d %d", $n, $x);
$a = [];
$a[0] = 1;
$p = [];
$p[0] = 1;
for ($i = 1; $i < 51; ++$i) {
    $a[$i] = $a[$i - 1] * 2 + 3;
    $p[$i] = $p[$i - 1] * 2 + 1;
}

function search($n, $x) {
    global $a, $p;
echo "n: $n, x: $x", PHP_EOL;
    if ($n == 0) {
        if ($x == 0) return 0;
        if ($x == 1) return 1;
    }

    if ($x == 1) return 0;
    if ($x < $a[$n - 1] + 2) return search($n - 1, $x - 1);
    if ($x == $a[$n - 1] + 2) return $p[$n - 1] + 1;
    if ($x < 2 * $a[$n - 1] + 3) return $p[$n - 1] + 1 + search($n - 1, $x - ($a[$n - 1] + 2));
    if ($x == 2 * $a[$n - 1] + 3) return 2 * $p[$n - 1] + 1;
}

echo search($n, $x);