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;
const long long N = 1e6 + 5, inf = 1e18, mod = 1e9 + 7;
const long double pi = 3.141592653589793238;
bool check_prime(long long n) {
  long long flag = 0, i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (n == 1) {
    return false;
  } else if (flag == 0 || n == 2 || n == 3) {
    return true;
  } else {
    return false;
  }
}
long long fastexpo(long long a, long long b, long long MOD) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result = (result * a) % MOD;
    }
    a = (a * a) % MOD;
    b = b / 2;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, i, p, c;
  cin >> n;
  set<long long> s;
  s.insert(n);
  c = 0;
  if (n == 1) {
    c++;
  }
  while (1) {
    n++;
    for (i = 1;; i++) {
      p = pow(10, i);
      if (n % p != 0) {
        p = p / 10;
        break;
      }
    }
    n /= p;
    s.insert(n);
    if (n == 1) {
      c++;
      if (c == 2) {
        break;
      }
    }
  }
  cout << ((long long)((s).size())) << "\n";
  return 0;
}
