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;

long long n,x;

long long dfs(long long n, long long x)
{
	if(n==0) {
		if(x > 0) return 1ll;  //如果x>0，左边都是B了 
			else return 0;  //之前吃光了 
	}
	if(x <= pow(2,n+1)-2) return dfs(n-1,x-1);   //汉堡包对称的，不到中间的话，递归n-1规模的子问题，最右边的B去掉，所以x-1 
		else return pow(2,n) + dfs(n-1,x - pow(2,n+1) + 1); //如果超过对称部分，则在n-1规模子问题中寻找剩下部分能吃多少p 
} 


int main()
{
	cin >> n >> x;
	cout << dfs(n,x);
	return 0;
}