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
import collections
n,x = map(int, raw_input().split(' '))
x-=1
mem = {}
def compute(n):
	if n in mem:
		return mem[n]
	if n == 0:
		return (1,1)
	else:
		h,p = compute(n-1)
		return (2*h + 3, 2*p + 1) 	
def iot(n,x,r,h):
	if h[0] > x: return
	if n == 0:
		h[0] += 1
		r[0] += 1
		return
	h[0] += 1
	if h[0] > x: return
	
	if h[0] + compute(n-1)[0] <= x:
		hh,pp = compute(n-1)
		h[0] += hh
		r[0] += pp
	else:
		iot(n-1, x, r, h)
	if h[0] > x: return
	
	h[0] += 1
	r[0] += 1

	if h[0] > x: return
	
	if h[0] + compute(n-1)[0] <= x:
		hh,pp = compute(n-1)
		h[0] += hh
		r[0] += pp
	else:
		iot(n-1, x, r, h)
	
	
	if h[0] > x: return
	
	h[0] +=1		
r,h = [0],[0]
iot(n,x,r,h)
print r[0]