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 long int ans = 0;
int sd(long long int rem) {
  int sum = 0;
  while (rem) {
    sum += rem % 10;
    rem /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int pos[13] = {0,           9,         99,         999,
                           9999,        99999,     999999,     9999999,
                           99999999,    999999999, 9999999999, 99999999999,
                           999999999999};
  long long int n;
  cin >> n;
  if (n >= pos[12]) {
    cout << 108 + sd(n - pos[12]) << "\n";
    return 0;
  }
  long long int rem;
  for (int i = 0; i < 12; i++) {
    if (n >= pos[i] && n < pos[i + 1]) {
      rem = n - pos[i];
      ans += 9 * (i);
      break;
    }
  }
  ans += sd(rem);
  cout << ans << "\n";
  return 0;
}