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;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
typedef long long ll;

vector<ll> a;
vector<ll> p;

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

int main(){
	ll n,x;
	cin >> n >> x;
	a.resize(n+1,0);
	p.resize(n+1,0);
	a[0] = 1;
	p[0] = 1;
	REP(i,n){
		a[i] = 2*a[i-1] + 3;
		p[i] = 2*p[i-1] + 1;
	}
	ll ans = solve(n, x);
	cout << ans << endl;
    return 0;
}