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)
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

vector<ll> lay(50+1), paty(50+1);

ll solve(ll n, ll x) {
	if(n == 0) return 0;
	if(x == 0) return 0;
	if(x < (lay[n]+1)/2) return solve(n-1, x-1);
	if(x == (lay[n]+1)/2) return 1LL + solve(n-1, lay[n-1]);
	if(x < lay[n]) return solve(n-1, x-lay[n-1]-2) + 1LL + solve(n-1, lay[n-1]);
	if(x == lay[n]) return paty[n];
}

int main() {
	ll n, x;
	cin >> n >> x;
	
	lay[0] = 1;
	paty[0] = 1;
	rep(i, n) {
		lay[i+1] = 2 * lay[i] + 3;
		// cout << lay[i+1] << endl;
	}
	rep(i, n) {
		paty[i+1] = 2 * paty[i] + 1;
		// cout << paty[i+1] << endl;
	}
	cout << solve(n, x) << endl;
	
	return 0;
}