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 ll = int_fast64_t;

ll thick[51], patty[51];

ll recur(ll n, ll x){
	if(n==0 && x) return 1;

	if(x<=1) return 0;
	if(x<=1+thick[n-1]) return recur(n-1, x-1);
	if(x<=2+thick[n-1]*2) return 1 + patty[n-1] + recur(n-1, x-2-thick[n-1]);
	return patty[n];
}

int main(){
	thick[0] = patty[0] = 1;
	for(int i=0; i<50; i++){
		thick[i+1] = thick[i] * 2 + 3;
		patty[i+1] = patty[i] * 2 + 1;
	}

	ll n, x;
	scanf("%ld %ld", &n, &x);
	printf("%ld\n", recur(n, x));
	return 0;
}
