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 rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

ll a[60], p[60];
ll f(ll n, ll x)
{
	if (n == 0) return 0;
	if (x == 1) return 0;
	else if (x <= a[n - 1] + 1)
		return f(n - 1, x - 1);
	else if (x <= a[n - 1] + 2)
		return p[n - 1] + 1; 
	else if (x <= a[n] - 1)
		return f(n - 1, x - a[n - 1] - 2) + p[n - 1] + 1;
	else if (x == a[n])
		return 2 * p[n - 1] + 1;
}

signed main(void)
{
	ll n, x, ans;

	cin >> n >> x;
	a[0] = 1;
	p[0] = 1;
	for(int i = 1; i <= n; i++)
	{
		a[i] += 2 * a[i - 1] + 3; 
		p[i] += 2 * p[i - 1] + 1; 
	}
	ans = f(n, x);
	cout << ans << endl;
	return 0;
}