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<iostream>
#include<string>
#include<vector>
#include<list>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<set>
#include<map>
#include<queue>
#include<cstring>
#include<sstream>
using namespace std;

typedef long long ll;

ll p[50],a[50];

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

int main() {
  int N; ll X; cin >> N >> X;
  p[0] = a[0] = 1;
  for ( int i = 1; i <= N; i++ ) {
    p[i] = 2*p[i-1]+1;
  }
  for ( int i = 1; i <= N; i++ ) {
    a[i] = 2*a[i-1]+3;
  }

  cout << func(N,X) << endl;
  return 0;
}