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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <array>

using namespace std;

// 厚さ
vector<long long> b(51, 0);
// パティの枚数
vector<long long> p(51, 0);

long long barger(int k){
    if(k == 0){
        b[0] = 1;
        return 1;
    }
    else {
        b[k] = 3 + 2 * barger(k-1);
        return b[k];
    }
}

long long patti(int k){
    if(k == 0){
        p[0] = 1;
        return 1;
    }
    else {
        p[k] = 1 + 2 * patti(k-1);
        return p[k];
    }
}

int main()
{
    long long n, x;
    cin >> n >> x;
    x--;
    long long max = barger(n);
    long long ans = patti(n);
    ans = p[n-1];
    long long left = 0;
    long long mid;
    long long right = max;
    /*
    for(int i=0; i<=50; ++i){
        cout << i << " " << b[i] << " " << p[i] << endl;
    }
    */
    //cout << max << " p -> " << ans << endl;
    for(int i=n; i>=0; --i){
        mid = (left + right) / 2;
        if(mid == x){
            ans++;
            break;
        }
        else if(x < mid){
            right = mid;
            left++;
            ans -= (i >= 2) ? p[i-2] : 0;
        }
        else {
            left = mid + 1;
            right--;
            ans += (i >= 2) ? p[i-2] : 0;
            right--;
        }
        // cout << i << ": " << left << " " << right << " " << ans << endl;
    }
    cout << ans << endl;
}