Problem: You are given a positive integer n.

Let S(x) be sum of digits in base 10 representation of x, for example, S(123) = 1 + 2 + 3 = 6, S(0) = 0.

Your task is to find two integers a, b, such that 0 ≤ a, b ≤ n, a + b = n and S(a) + S(b) is the largest possible among all such pairs.

Input

The only line of input contains an integer n (1 ≤ n ≤ 10^{12}).

Output

Print largest S(a) + S(b) among all pairs of integers a, b, such that 0 ≤ a, b ≤ n and a + b = n.

Examples

Input

35


Output

17


Input

10000000000


Output

91

Note

In the first example, you can choose, for example, a = 17 and b = 18, so that S(17) + S(18) = 1 + 7 + 1 + 8 = 17. It can be shown that it is impossible to get a larger answer.

In the second test example, you can choose, for example, a = 5000000001 and b = 4999999999, with S(5000000001) + S(4999999999) = 91. It can be shown that it is impossible to get a larger answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second)
    return a.first > b.first;
  else
    return (a.second > b.second);
}
long long int fact(long long int n) {
  if (n == 0 || n == 1)
    return 1;
  else
    return n * fact(n - 1);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, st;
  long long int x = 0, temp = 0, ans = 0;
  cin >> s;
  if (s.size() == 1) {
    cout << s << endl;
    return 0;
  }
  for (int i = 1; i < s.size(); i++) st += '9';
  stringstream ss(s);
  ss >> x;
  stringstream sm(st);
  sm >> temp;
  long long int te = x - temp;
  while (te) {
    ans += (te % 10);
    te /= 10;
  }
  cout << ans + (s.size() - 1) * 9 << endl;
}