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: Wrong Answer)

<?php

//input
fscanf(STDIN,"%d %d",$n,$eat);

$atusa = [1];
$paty = [1];
for($i=1;$i<=$n;$i++){
    $paty[$i] = $paty[$i-1]*2+1;
    $atusa[$i] = 3+$atusa[$i-1]*2;
}

echo EatB($n,$eat,$paty,$atusa);

function EatB($n,$eat,$paty,$atusa){
    //echo $n."バーガーを".$eat."食う処理を開始中\n";
    if($eat==1)return 0;
    $eat--;//最初のバーガー食う
    
    if($eat==$atusa[$n-1]){//ちょうどバーガー食う
        //echo "paty".$paty[$n-1]."をちょうど食べました1 - ".($n)."バーガー\n";
        return $paty[$n-1];
    }elseif($eat<$atusa[$n-1]){
        //echo "2 : 食べかけ開始\n";
        $an = EatB($n-1,$eat,$paty,$atusa);
        //echo "2 : 食べかけ終了\n";
        //echo "paty".$an."を食べかけました - ".($n)."バーガー\n";
        return $an;
    }elseif($eat==$atusa[$n-1]+1){//真ん中の食う
        //echo "paty".$paty[$n-1]."+1をちょうど食べました3 - ".($n-1)."バーガー\n";
        return $paty[$n-1]+1;
    }else{//端の食う
        //echo "4 : 食べかけ開始\n";
        $an = $paty[$n-1]+1 + EatB($n-1,$eat-($atusa[$n-1]+1),$paty,$atusa);
        //echo "4 : 食べかけ終了\n";
        //echo "paty".$an."を食べかけました - ".($n)."バーガー\n";
        return $an;
    }
}
/*

B BPPPB P BPPPB B


2バーガーを3食う処理を開始中
2 : 食べかけ開始
1バーガーを2食う処理を開始中
paty1をちょうど食べました1 - 1バーガー
2 : 食べかけ終了
paty4を食べかけました - 2バーガー
4


*/

?>