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 vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
#define tll tuple<ll, ll, ll>
#define all(c) c.begin(),c.end()

const ll inf = 1000000001;
const ll INF = 1e18;
const ll MOD = 1000000007;

ll h[50];
ll p[50];

ll calc(ll n, ll x){
  if (n==0){
    if (x <= 0) return 0;
    else return 1;
  }
  if (x == 1) return 0;
  if (1 < x && x <= h[n-1]+1) return calc(n-1, x-1);
  if (x == h[n-1]+2) return p[n-1] + 1;
  if (h[n-1]+2 < x && x <= 2*h[n-1]+2) return p[n-1] + 1 + calc(n-1, x-h[n-1]-2);
  if (x == 2*h[n-1]+3) return 2*p[n-1] + 1;
  //if (n==0){
  //  if (x <= 0) return 0;
  //  else return 1;
  //}
  //else if (x <= h[n-1]+1) return calc(n-1, x-1);
  //else return p[n-1] + 1 + calc(n-1, x-h[n-1]-2);
}

int main() {
  ll n, x;
  cin >> n >> x;
  h[0] = 1;
  p[0] = 1;
  for(int i=1;i<=50;i++){
    h[i] = 2*h[i-1] + 3;
    p[i] = 2*p[i-1] + 1;
  }
  
  ll ans = calc(n,x);
  cout << ans << endl;
}