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 <string>
#include <vector>
#include <algorithm>

#define rep(i,s,n)for(int i = s;i<n;i++)
#define repe(i,s,n)for(int i = s;i<=n;i++)
#define repe_r(i,s,n)for(int i = n;i>=0;i--)
using namespace std;
typedef long long ll;
static const ll MOD = 1000000007;// 1e9 + 7
ll Pow(ll n, ll p) {
	if (p == 0)return 1;
	if (p == 1)return n;
	ll t = Pow(n, p / 2);
	if (p & 1) return (t * t % MOD) * n % MOD;
	else return t * t % MOD;
}
int main() {
	ll n, x; cin >> n >> x;
	ll a = Pow(2, n + 2) - 3;

	ll ans = 0;

	repe_r(i, 0, n) {
		//if (i == 0) {
		//	ans++;
		//	break;
		//}

		if (x > a / 2 + 1) {
			ans += Pow(2, i);
			x -= Pow(2, i + 1) - 1;
		}
		else if (x < a / 2 + 1) {
			x--;
		}
		else {
			ans += Pow(2, i);
			break;
		}
		a = Pow(2, i + 1) - 3;
	}

	cout << ans << endl;

	return 0;
}