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>
#define LL long long
using namespace std;
LL n, m, i, j, k, a[55], c[55];
LL dfs(LL n, LL k){
	if(k <= 0 || n < 0) return 0;
	if(k >= c[n]) return a[n];
	LL s = dfs(n-1, k-1);
	if(k > c[n-1]+1) s++;
	s += dfs(n-1, k-c[n-1]-2);
	return s;
}
int main(){
	scanf("%lld%lld", &n, &k);
	a[0] = c[0] = 1;
	for(i=1; i<=n; i++){
		a[i] = a[i-1] * 2 + 1;
		c[i] = c[i-1] * 2 + 3;
	}
	printf("%lld\n", dfs(n, k));
	return 0;
}