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>
const int MAX = (int)(2e5) + 7;
const int MOD = (int)(1e9) + 7;
const int INF_I = (int)(1e9) + 7;
const long long INF_LL = (long long)(1e18) + 7;
using namespace std;
long long n;
template <typename T>
void info(T& a) {
  for (auto& el : a) cout << el << ' ';
  cout << '\n';
}
template <typename T>
bool chkmax(T& a, T b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chkmin(T& a, T b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
void boot() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
void test() {}
long long f(long long x) {
  x += 1LL;
  while (x % 10 == 0) x /= 10;
  return x;
}
void solve() {
  test();
  set<long long> s;
  do {
    s.insert(n);
    n = f(n);
  } while (!s.count(n));
  cout << (int)((s).size()) << '\n';
}
signed main() {
  boot();
  while (cin >> n) solve();
}
