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: Accepted)

#include<bits/stdc++.h>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
#define rep(i,n) for(ll i = 0;i < (ll)n;i++)
#define ALL(x) (x).begin(),(x).end()
#define MOD 1000000007

vector<ll> len(60,1),p(60,1);

ll dfs(ll n,ll x){
  if(n == 0)return x ? 1 : 0;
  if(x == 1)return 0;
  else if(x <= len[n-1]+1)return dfs(n-1,x-1);
  else if(x == len[n-1]+2)return p[n-1]+1;
  else if(x <= 2*len[n-1]+2)return p[n-1]+1+dfs(n-1,x-len[n-1]-2);
  else return p[n-1]*2+1;
}

int main(){
  
  rep(i,55)len[i+1] = 2*len[i]+3;
  rep(i,55)p[i+1] = 2*p[i]+1;
  ll n,x;
  cin >> n >> x;
  cout << dfs(n,x) << "\n";
  





  return 0;
}