def is_woodall(x):
	 
	if (x % 2 == 0): 
		return False
	if (x == 1): 
		return True
	x = x + 1 
	p = 0
	while (x % 2 == 0): 
		x = x/2
		p = p + 1
		if (p == x): 
			return True
	return False
def count_ways(n):
	 
	A = [0] * (n + 1) 
	B = [0] * (n + 1) 
	A[0] = 1
	A[1] = 0
	B[0] = 0
	B[1] = 1
	for i in range(2, n+1): 
		A[i] = A[i - 2] + 2 * B[i - 1] 
		B[i] = A[i - 1] + B[i - 2] 
	return A[n] 