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>
using namespace std;
#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define ALL(a) a.begin(), a.end()
ll dx[4] = { 0, 1, 0, -1 };
ll dy[4] = { 1, 0, -1, 0 };
const ll INF = 1e14;
const ll MOD = 1e9 + 7;
// level iバーガーの厚さ a[i] := 2*a[i-1] + 3, a[0] = 1
// level iバーガーのパティの総数 p[i] := 2*p[i-1] + 1, p[0] = 1
vector<ll> a = { 1 }, p = { 1 };


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

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ll n, x;
    cin >> n >> x;
    for (int i = 0; i < n; i++) {
        a.push_back(a[i] * 2 + 3);
        p.push_back(p[i] * 2 + 1);
    }
    cout << f(n, x) << endl;
    return 0;
}