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;
void debug(int n, ...) {
  va_list v1;
  va_start(v1, n);
  for (int i = 0; i < n; i++) {
    long long int val = va_arg(v1, int);
    printf("%lld ", val);
  }
  va_end(v1);
  cout << endl;
}
long long int f(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
long long int bs(long long int n) {
  long long int ans = 0;
  while (n) {
    ans++;
    n /= 10;
  }
  return ans;
}
int main() {
  long long int n;
  cin >> n;
  long long int basa = bs(n);
  basa--;
  long long int carp = 1;
  for (long long int i = 0; i < basa; i++) carp *= 10;
  long long int mod = n % carp;
  long long int ans = f(mod + 1) + f(n - mod - 1);
  cout << ans;
}