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;
using ll = long long;
using ull = unsigned long long;

#ifndef __MACRO_H__
#define __MACRO_H__

#define all(collection) (collection).begin(), (collection).end()  // begin to end
#define rep(i, begin, end) for (ll i = begin; i < end; i++)       // repeat
#define repr(i, begin, end) for (ll i = begin; end < i; i--)      // repeat reverse
#define size(collection) ((ll) (collection).size())               // collection size

#endif

vector<ull> lmax(50);
vector<ull> pmax(50);

void Solve(ll lv, ll& x, ll& ans)
{
	x--;
	if (x <= 0 || lv == 0)
		return;

	if (x <= lmax[lv - 1])
	{
		Solve(lv - 1, x, ans);
		return;
	}

	x -= lmax[lv - 1];
	ans += pmax[lv - 1];
	if (x <= 0)
		return;

	x--; ans++;
	if (x <= 0)
		return;

	if (x <= lmax[lv - 1])
	{
		Solve(lv - 1, x, ans);
		return;
	}

	x -= lmax[lv - 1];
	ans += pmax[lv - 1];
	if (x <= 0)
		return;

	x--;
}

int main(void)
{
	ll n, x; cin >> n >> x;

	lmax[0] = pmax[0] = 1;
	rep(i, 1, 50)
	{
		lmax[i] = lmax[i - 1] * 2 + 3;
		pmax[i] = pmax[i - 1] * 2 + 1;
	}

	ll ans = 0;
	Solve(n, x, ans);

	cout << ans;
	return 0;
}