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 JavaScript (node.js v5.12) (judge verdict: Accepted)

var lines = []; var f;
var readline = require('readline');

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    lines.push(x);
});

rl.on('close', function () {

    /* --TLE-- */

    var N = Number(lines[0].split(" ")[0]);
    var X = Number(lines[0].split(" ")[1]);

    var result = 0;

    // バーガーの厚さの数列
    // an+1 = 2an + 3
    // (an+1 + 3) = 2(an +3)
    // 2(an + 3) = 2^n(a1 + 3)
    // an =  2^(n+2) - 3

    // パティの量
    // bn+1 = 2bn + 1
    // bn+1 + 1 = 2(bn + 1)
    // 2(bn + 1) = 2^n(b1 + 1)
    // bn = 2^(n+1) - 1

    console.log(dfs(N, X));
});

// 第nバーガー， 第x層までルンルンが食べる
function dfs(n, x) {
    var len = Math.pow(2, n + 1) - 3; // バーガーの厚さ
    var num = Math.pow(2, n) - 1; // パティの厚さ

    if (n === 0) { return (1); } // 0段バーガーのパティは1枚

    if (x === 1) { return (0); } // ルンルンが1層しか食べない場合，それはパン
    else if (x <= len + 1) { return (dfs(n - 1, x - 1)) } // ルンルンが食べるハンバーガーが半分以下ならば
    else if (x == len + 2) { return (num + 1) }// ルンルンが食べるハンバーガーがちょうど半分
    else if (x <= (len + 1) * 2) { return (num + 1 + dfs(n - 1, x - len - 2)); }　// ルンルンが食べるハンバーガーが半分以上全部未満
    else { return (num * 2 + 1) }// ルンルンがハンバーガーを全部食べる
    
    // dfsの引数にx-1を渡すのは，最下層がパンであるため
}