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 <iostream>
  #include <iomanip>
  #include <algorithm>
  #include <cmath>
  #include <queue>
  #include <list>
  #include <limits>
  #include <string>
  #include <vector>
  #include <stack>
  using namespace std;
  typedef long long ll;

  typedef pair<ll, ll> curr;
  const int MOD = 998244353;


  int gcd(int a, int b){
    if(b == 0) return a;
    return gcd(b, a%b);
  }

  ll calc(int a, int b, int p){
    if(b == 0) return 1;
    if(b % 2 == 0){
      ll d = calc(a, b/2, p);
      return (d*d)%p;
    }else{
      return (a*calc(a, b-1, p))%p;
    }

  }
  ll a[50];
  ll p[50];
  ll f(ll n, ll x){
    if(n == 0) return 0;
    if(x <= 1) return 1;
    if(x <= a[n-1]+1) return f(n-1,x-1);
    if(x == 2 + a[n-1]) return p[n-1]+1;
    if(x && x <= 2+2*a[n-1]) return p[n-1]+1+f(n-1 , x-2-a[n-1]);
    if(x == 3 + 2*a[n-1]) return 2*p[n-1]+1;
  }

  int main(){
    ll N,X;
    cin >> N >> X;
    a[0] = 1;
    for(int i=1;i<N;i++) a[i] = 2*a[i-1] + 3;
    p[0] = 1;
    for(int i=1;i<N;i++) p[i] = 2*p[i-1] + 1;
    cout <<  f(N,X);
  }