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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

int N;
ll X;
vector<ll> nAll;
vector<ll> nP;

ll f(int l, ll x) {
  if(x == 0) return 0;
  if(l == 0) return x;

  if(x == nAll[l]) return nP[l];
  if(x < nAll[l-1] + 2) return f(l-1, x-1);
  return 1 + nP[l-1] + f(l-1, x - nAll[l-1] - 2);
}

int main () {
  cin >> N >> X;

  nAll = vector<ll>(N+1, 1);
  nP = vector<ll>(N+1, 1);
  rep(i,N) {
    nAll[i+1] = 2*nAll[i] + 3;
    nP[i+1] = 2*nP[i] + 1;
  }

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