def kaibun(y):
	x = len(str(y))
	flag = 0
	X = [9,9]
	i = 2
	while x > i:
		if i % 2 == 0:
			X.append(X[i - 1] * 10)
		else:
			X.append(X[i - 1])
		i = i + 1
	a = sum(X[:x - 1])
	for j in range(1, x // 2 + 1):
		if j == 1:
			a = a + (int(str(y)[j - 1]) - 1) * X[x - 2 * j + 1]//9
		else:
			a = a + (int(str(y)[j - 1])) * X[x - 2 * j + 1] // 9
	if str(y)[:x//2][::-1] and str(y)[x//2 + 1:] and int(str(y)[:x//2][::-1]) <= int(str(y)[x//2:]) and x % 2 == 0:
		a = a + 1
		flag = 1
	elif x % 2 == 1:
		a = a + int(str(y)[x//2])
		if str(y)[:x//2][::-1] and str(y)[x//2 + 1:] and int(str(y)[:x//2][::-1]) <= int(str(y)[x//2 + 1:]):
			a = a + 1
			flag = 1
	if x == 1:
		return y,flag
	return a,flag
	# X[i] は i 桁の自然数のうち回文となるものの個数
n = input()
if len(n) < 10:
	print(0)
else:
	ni = n[:-9]
	k = int(ni)
	answer,flag = kaibun(k)
	if len(n) == 10 and int(n) % 10**9 < int(n[0]):
		answer = answer - 1
	if flag == 1 and int(ni[::-1]) > int(n[-9:]):
		answer = answer - 1
	print(answer)