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;
void read(int &x) {
  x = 0;
  char ch = getchar();
  int pd = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      pd = -pd;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= pd;
}
void write(const int &x) {
  char ggg[10001];
  int s = 0;
  int tmp = x;
  if (tmp == 0) {
    putchar('0');
    return;
  }
  if (tmp < 0) {
    tmp = -tmp;
    putchar('-');
  }
  while (tmp > 0) {
    ggg[s++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (s > 0) {
    putchar(ggg[--s]);
  }
}
map<int, bool> vis;
int n, a[10001], len;
int main() {
  read(n);
  int tmp = n;
  int ans = 0;
  while (!vis[tmp]) {
    vis[tmp] = 1;
    ans++;
    tmp++;
    while (tmp % 10 == 0) tmp /= 10;
  }
  write(ans);
}
