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.

Examples

Input

2 7


Output

4


Input

1 1


Output

0


Input

50 4321098765432109


Output

2160549382716056
#include<bits/stdc++.h>
using namespace std;
typedef long long lint;

lint A[55], P[55];

lint rec(lint n, lint x){
    if(n==0){
        if(x<=0) return 0LL;
        return 1LL;
    }

    if(x <= 1 + A[n-1]) return rec(n-1, x-1);
    return P[n-1] + 1 + rec(n-1, min(A[n-1], x-A[n-1]-2)); 
}

int main(){
    lint n, x; cin >> n >> x;

    A[0] = P[0] = 1;
    for(int i=1;i<=n;i++){
        A[i] = 2 * A[i-1] + 3;
        P[i] = 2 * P[i-1] + 1;
    }

    cout << rec(n, x) << endl;
    return 0;
}