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 ll;

/*
L(0) := P
L(n) := B L(n-1) P L(n-1) B

f(n) := L(n) の文字数
g(n) := L(n) のPの個数
*/

ll f(int n){ return n==0 ? 1 : f(n-1)*2+3; }
ll g(int n){ return n==0 ? 1 : g(n-1)*2+1; }

ll calc(int n, ll x){
    if(x<=0) return 0;
    if(n==0) return 1;
    if(x > f(n)/2) return g(n-1) + 1 + calc(n-1, x - (f(n)/2+1));
    else return calc(n-1, x-1);
}

int main(void){
    int n;
    ll x;
    cin >> n >> x;
    cout << calc(n, x) << endl;
    return 0;
}
