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 <algorithm>
#include <vector>
#include <string>
#include <numeric>
#include <map>
#include <iomanip>
#include <limits.h>

using namespace std;    
typedef long long ll;
typedef pair<ll, ll> pll;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const bool DEBAG = false;

ll bans[51]; // レベルNバーガーの長さ
ll pat[51]; // レベルNバーガーにパティが何枚あるか


// levelバーガーをしたからx枚のとき，何枚のパティを食べれるか
ll eat(ll L, ll x){

    if(x <= 1){
        return 0;
    }
    else if(x <= 1 + bans[L-1]){
        return eat(L - 1, x - 1);
    }
    else if(x == 2 + bans[L - 1]){
        return eat(L - 1, x - 1) + 1;
    }
    else {
        return pat[L - 1] + eat(L - 1, x - 1 - bans[L - 1] - 1) + 1;
    }

}

int main(){

    ll N, X;
    cin >> N >> X;

    bans[0] = 1;
    for(int i = 1; i <= N; i++){
        bans[i] = 2 * bans[i-1] + 3;
    }
    

    pat[0] = 1;
    for(int i = 1; i <= N; i++){
        pat[i] = 2 * pat[i-1] + 1;
    }

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


}