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.
#include <bits/stdc++.h>
using namespace std;
string s, s1;
int a[1000];
set<string> l;
void fun() {
  memset(a, 0, sizeof(a));
  int len = s.length();
  int j = 0;
  for (int i = len - 1; i >= 0; i--) {
    a[j++] = s[i] - '0';
  }
  a[0]++;
  for (int i = 0; i < len; i++) {
    if (a[i] >= 10) {
      a[i + 1] += a[i] / 10;
      a[i] %= 10;
    }
  }
  if (a[len] != 0) len++;
  int index;
  for (int i = 0; i < len; i++) {
    if (a[i] == 0) continue;
    index = i;
    break;
  }
  for (int i = len - 1; i >= index; i--) {
    s1 += a[i] + '0';
  }
}
int main() {
  while (cin >> s) {
    l.clear();
    l.insert(s);
    s1 = "";
    fun();
    int sum = 1;
    while (l.count(s1) != 1) {
      l.insert(s1);
      s = s1;
      s1 = "";
      fun();
      sum++;
    }
    cout << sum << endl;
  }
  return 0;
}
