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 <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
ll n, x;

ll fast_pow(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if ((b & 0x1))
            ans *= a;
        a *= a;
        b >>= 1;
    }
    return ans;
}

ll get_num(ll total, ll x) {
    if (x == total) {
        return (total + 1ll) / 2ll;
    }
    if (x == 1) {
        return 0;
    }
    x--;
    ll tt = (total - 3ll) / 2ll;
    if (x < tt) return get_num(tt, x);
    else if (x == tt) return get_num(tt, x);
    else if (x == tt + 1) return 1 + get_num(tt, x - 1);
    else {
        ll ans1 = (tt + 1ll) / 2ll;
        x -= tt;
        return ans1 + 1 + get_num(tt, x - 1);
    }
    return 0;
}
int main() {
    cin >> n >> x;
    ll total = 4ll * fast_pow(2, n) - 3;
    ll ans = get_num(total , x);
    cout << ans << endl;
}