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: Accepted)

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long

using namespace std;

const ll MAXN = 51;
ll memo[MAXN];

ll powfunc(ll a, ll n) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}

ll dfs(ll n, ll x) {
	if (n == 0){
		return memo[0] = 1;
	}
	if (memo[n] != 0) {
		return memo[n];
	}

	//A[n-1]の長さを求める
	ll alen = powfunc(2, n+1) - 3;
	if (x > 2 * (alen + 2)){
		return memo[n] = 2 * dfs(n-1, alen) + 1;
	} else if (x > alen + 2) {
		return dfs(n-1, x - (alen + 2)) + 1 + dfs(n-1, alen);
	} else if (x > alen + 1) {
		return 1 + dfs(n-1, alen);
	} else if (x > 1) {
		return dfs(n-1, x-1);
	} else {
		return 0;
	}
}

int main() {
	ll N, X;
	cin >> N >> X;

	ll ans = dfs(N, X);

	cout << ans << "\n";

	return 0;
}