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 <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <cstring>

using namespace std;
// ascending order
#define vsort(v) sort(v.begin(), v.end())
// descending order
#define vsort_r(v) sort(v.begin(), v.end(), greater<int>())
#define vunique(v) unique(v.begin(), v.end())
#define mp make_pair
#define ts(x) to_string(x)
#define rep(i, a, b) for(int i = (int)a; i < (int)b; i++)
#define repm(i, a, b) for(int i = (int)a; i > (int)b; i--)
#define bit(a) bitset<8>(a)
#define des_priority_queue priority_queue<int, vector<int>, greater<int> >
#define all(v) (v).begin(), (v).end()
typedef long long ll;
typedef pair<int, int> P;
const ll INF = 1e18;

ll N, X;
ll a[51], p[51];
ll F(ll N, ll X) {
    if(N == 0) {
        if(X <= 0) return 0;
        return 1;
    }
    if(X == 1) return 0;
    if(X <= 1 + a[N - 1]) return F(N - 1, X - 1);
    if(X <= 2 * a[N - 1] + 2) return p[N - 1] + 1 + F(N - 1, X - 2 - a[N - 1]);
    return 2 * p[N - 1] + 1;


}
int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

    cin >> N >> X;
    a[0] = 1;
    p[0] = 1;
    rep(i, 1, N + 1) {
        a[i] = 2 * a[i - 1] + 3;
        p[i] = 2 * p[i - 1] + 1;
    }

    cout << F(N, X) << endl;

}