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 <iostream>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
using namespace std;
const int INF=1000000007;

int main (void){
    long long int n,x,i,ps=0;
    cin >> n >> x;
    for(i=n;i>=0;i--){
    	if(x==0) break;
    	if(i==0){
    		ps++;
    		break;
    	}
    	if(x==pow(2,i+2)-3) x=x-1;
    	if(x>=pow(2,i+1)-1){
    		ps+=pow(2,i);
    		x=x-pow(2,i+1)+1;
    	} else {
    		x=x-1;
    	}
    }
    cout << ps << endl;
    return 0;
}