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
//Problem D
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int n;ll k,a[55],b[55];
ll solve(int n,ll k){//printf("%d %lld\n",n,k);
	if(!n||!k)return k;
	if(k<=a[n-1]+1)return solve(n-1,k-1);
	return b[n-1]+1+solve(n-1,k-a[n-1]-2-(k==a[n]));
}
int main(){
    scanf("%d%lld",&n,&k);
	a[0]=b[0]=1;for(int i=1;i<=n;i++)a[i]=3+a[i-1]*2,b[i]=1+b[i-1]*2;
	printf("%lld",solve(n,k));
    return 0;
}