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>
#define FOR(i, k, n) for(int i = (k); i < (n); i++)
#define FORe(i, k, n) for(int i = (k); i <= (n); i++)
#define FORr(i, k, n) for(int i = (k)-1; i > (n); i--)
#define FORre(i, k, n) for(int i = (k)-1; i >= (n); i--)
#define REP(i, n) FOR(i, 0, n)
#define REPr(i, n) FORre(i, n, 0)
#define ALL(x) (x).begin(), (x).end()
#define ALLr(x) (x).rbegin(), (x).rend()
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
using namespace std;
using ll = long long;

const int INF = 1001001001;

vector<ll> bg(51), p(51);

ll f(ll n, ll x){
  if(x == 1) return 0;
  if(1 < x && x <= bg[n-1]+1) return f(n-1, x-1);
  if(x == bg[n-1]+2) return p[n-1]+1;
  if(bg[n-1]+2 < x && x <= 2*bg[n-1]+2) return p[n-1]+1 + f(n-1, x-bg[n-1]-2);
  if(x == bg[n]) return p[n];
}

int main(void){
  ll n, x;
  cin >> n >> x;

  bg[0] = 1;
  p[0] = 1;
  REP(i, n){
    bg[i+1] = 2*bg[i]+3;
    p[i+1] = 2*p[i]+1;
  }
  cout << f(n, x) << endl;
  return 0;
}