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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <utility>
#include <algorithm>
#include <bitset>
#include <numeric>
#include <cmath>
#include <tuple>
#include <queue>
#include <set>

using namespace std;

typedef long long ll;
const ll INF = 1LL<<60;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }


ll a[51] = {0};
ll p[51] = {0};

ll get_ans(int N, ll X){
    if(X == 1) return 0;
    if(1 < X && X <= (1 + a[N-1])) return get_ans(N-1, X-1);
    if(X == (2 + a[N-1])) return p[N-1] + 1;
    if(X > (2 + a[N-1]) && X <= (2 + 2 * a[N-1])){
        return p[N-1] + 1 + get_ans(N-1, X-2-a[N-1]);
    }
    else{
        return 2 * p[N-1] + 1;
    }
}

int main(){
    cin.tie(0);
   	ios::sync_with_stdio(false);
    
    int N;
    ll X;

    cin >> N >> X;

    a[0] = 1;
    p[0] = 1;
    for(int i = 1; i < 51; i++){
        a[i] = 2 * a[i-1] + 3;
        p[i] = 2 * p[i-1] + 1;
    }

    cout << get_ans(N, X) << endl;

    return 0;
}