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
import bisect
import heapq
from collections import defaultdict


def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, x, y = egcd(b % a, a)
        return (g, y - (b // a) * x, x)


def mulinv(b, n):
    g, x, _ = egcd(b, n)
    if g == 1:
        return x % n


primes = []


def isprime(n):
    for d in range(2, int(math.sqrt(n))+1):
        if n % d == 0:
            return False
    return True


def argsort(ls):
    return sorted(range(len(ls)), key=ls.__getitem__)


def f(p=0):
    if p == 1:
        return map(int, input().split())
    elif p == 2:
        return list(map(int, input().split()))
    else:
        return int(input())

n = f()

count = 1

st = True
mx = -1

e = 0
while(n>=10):
    n+=1
    while(n%10==0):
        n = n//10

    count+=1

print(count+8)