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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include<iostream>
using namespace std;
string barger[51];
long long p[51];
long long rinrin(int n, long long x) {
    if (x == 1) { //if n is not 0, last is B, and n is not 0
        return 0;
    } else if (x > 1 && x <= 1 + barger[n - 1].length() - 1) { //1 +'s 1 is first B
        return rinrin(n - 1, x - 1);
    } else if (x == 2 + barger[n - 1].length() - 1) { //2 +'s 2 are first B and center P
        return p[n - 1] + 1;
    } else if (2 + barger[n - 1].length() - 1 < x && x <= 2 + 2 * (barger[n - 1].length() - 1)) {
        return p[n - 1] + 1 + rinrin(n - 1, x - 1 - (barger[n - 1].length() - 1 + 1)); //the rest
    } else /*if (x == 3 + 2 * barger[n - 1].length() - 1)*/ {
        return 2 * p[n - 1] + 1;
    }
}
int main() {
    barger[0] = "P";
    p[0] = 1; //firstly p
    int n;
    long long x; // ~9*10^8
    cin >> n >> x;
    for (int i = 1; i <= n; i++) {
        barger[i] = "B" + barger[i - 1] + "P" + barger[i - 1] + "B";
        p[i] = 2 * p[i - 1] + 1;
    }
    cout << rinrin(n, x) << endl;
}