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>
#define rep(i, n) for (int i = 0; i < (n); i++)
typedef long long ll;
using namespace std;
 
vector<ll> cnt;
vector<ll> pat;
 
ll rec(ll N,ll X){
  if(X==0) return 0;
  if(N==0&&X==1) return 1;
  X--;
  if (X <= cnt[N - 1]) return rec(N - 1, X);
  X-=cnt[N-1];
  if (X == 1) return pat[N-1]+1LL;
  X--;
  if (X <= cnt[N - 1]) return pat[N-1] + rec(N - 1, X) + 1;
  return pat[N];
}
 
int main() {
    ll N,X;
    cin >> N >> X;
    cnt.resize(N+1); cnt[0] = 1;
    pat.resize(N+1); pat[0] = 1;
    rep(i,N){
        cnt[i + 1] = 3 + 2 * cnt[i];
        pat[i + 1] = 1 + 2 * pat[i];
    }
    cout << rec(N,X) << endl;
}