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 C# (.NET Core 3.1.201) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using static System.Math;

class MainClass
{
    static void Main()
    {
        var NX = ReadLine().Split();
        int N = int.Parse(NX[0]);
        long X = long.Parse(NX[1]);

        var all = new long[N + 1];
        all[0] = 1;
        var Patty = new long[N + 1];
        Patty[0] = 1;
        for (int i = 0; i < N; i++)
        {
            all[i + 1] = all[i] * 2 + 3;
            Patty[i + 1] = Patty[i] * 2 + 1;
        }
        int n = N;
        long ans = 0;
        while (n > 1)
        {
            if (X >= all[n] / 2+1)
            {
                ans += Patty[n] / 2+1;
                X -= all[n] / 2;
            }
            else if (X == all[n] / 2+1)
            {
                ans += Patty[n] / 2+1;
                break;
            }
            else if (X == all[n] / 2) {
                ans += Patty[n] / 2;
                break;
            }
            X--;
            n--;
        }
        
        if (n == 1&&X>0)
        {
            
            string s = "BPPPB".Substring(0, (int)X);
            ans += s.Count(_ => _ == 'P');
        }
        WriteLine(ans);
        
    }
}