Let's denote a function f(x) in such a way: we add 1 to x, then, while there is at least one trailing zero in the resulting number, we remove that zero. For example, 

  * f(599) = 6: 599 + 1 = 600 → 60 → 6; 
  * f(7) = 8: 7 + 1 = 8; 
  * f(9) = 1: 9 + 1 = 10 → 1; 
  * f(10099) = 101: 10099 + 1 = 10100 → 1010 → 101. 



We say that some number y is reachable from x if we can apply function f to x some (possibly zero) times so that we get y as a result. For example, 102 is reachable from 10098 because f(f(f(10098))) = f(f(10099)) = f(101) = 102; and any number is reachable from itself.

You are given a number n; your task is to count how many different numbers are reachable from n.

Input

The first line contains one integer n (1 ≤ n ≤ 10^9).

Output

Print one integer: the number of different numbers that are reachable from n.

Examples

Input


1098


Output


20


Input


10


Output


19

Note

The numbers that are reachable from 1098 are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1098, 1099.
import math
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b


def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c


def dv():
	n, m = map(int, input().split())
	return n,m


def dva():
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b


def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c


def dvs():
	n = int(input())
	m = int(input())
	return n, m


def pr(x):
	used = set()
	used.add(x)
	k = 10 - x % 10
	for i in range(x, x + k):
		if i % 10 != 0:
			used.add(i)
	x += k
	if x % 10 != 0:
		used.add(x)
	while x > 1:
		knn = 0
		while x % 10 == 0:
			x //= 10
		kn = 10 - x % 10
		if x == 1:
			for i in range(1, 10):
				used.add(i)
			used.add(x)
			return len(used)
		for i in range(x, x + kn):
			if i % 10 != 0:
				used.add(i)
			else:
				break
		x += kn
		k += kn + knn
	return len(used)
n = int(input())
print(pr(n))
