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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,ans=0;
int main(){
    cin>>n>>m;
    vector<ll> sumP(n+1),sumLayer(n+1);
    sumP.at(0)=1,sumLayer.at(0)=1;
    for(int i=1;i<n+1;i++){
        sumP.at(i)=2*sumP.at(i-1)+1;
        sumLayer.at(i)=2*sumLayer.at(i-1)+3;
    }
    while(n){
        if(m*2+1==sumLayer.at(n)){
            ans+=sumP.at(n-1)+1;
            m=0;
            break;
        }
        else if(m*2+1<sumLayer.at(n)){
            m--;
            n--;
        }
        else{
            ans+=sumP.at(n-1)+1;
            m-=(sumLayer.at(n-1)+2);
            n--;
        }
    }
    cout<<ans<<endl;
}