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;

using ll = long long;
vector<ll> A;
vector<ll> P;

ll dfs(int n, ll x) {
  if (n == 0) return 0;
  ll res = 0;
  if (x == 0) return res;
  res++; x--;
  if (x <= A[n-1]) return res + dfs(n-1, x);
  res += P[n-1] + 1; x -= A[n-1] + 1;
  if (x <= A[n-1]) return res + dfs(n-1, x);
  return P[n];
}

int main() {
  int N;
  ll X;
  cin >> N >> X;
  A.resize(N+1, 1);
  P.resize(N+1, 0);
  for (int i = 1; i < N+1; i++) {
    A[i] = A[i-1] * 2 + 3;
    P[i] = P[i-1] * 2 + 3;
  }
  cout << dfs(N, X) << endl;
}