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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

ll dp[51], dp2[51];

ll solve(int level, ll rest) {
    if (rest == 0) return 0;
    if (level == 0) return 1;

    if (rest <= 1) return 0;
    ll hoge = min(rest, dp[level - 1]);
    ll res = (hoge == dp[level - 1] ? dp2[level - 1] : solve(level - 1, hoge));
    rest -= hoge;

    if (rest == 0) return res;

    rest--;
    res++;

    hoge = min(rest, dp[level - 1]);
    res += (hoge == dp[level - 1] ? dp2[level - 1] : solve(level - 1, hoge));
    rest -= hoge;

    if (rest == 0) return res;
    rest--;
    return res;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    ll x;
    cin >> n >> x;

    dp[0] = 1;
    dp2[0] = 1;
    for (int i = 1; i <= n; i++) {
        dp[i] = 2 * dp[i - 1] + 3;
        dp2[i] = 2 * dp2[i - 1] + 1;
    }
    cout << solve(n, x) << endl;
    return 0;
}