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;
int sum(long long n) {
  int res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
int solve_slow(long long n) {
  int res = 0;
  for (long long a = 0, b = n; a <= b; a++, b--) {
    res = max(res, sum(a) + sum(b));
  }
  return res;
}
int solve_fast(long long n) {
  if (n <= 10) return (int)n;
  long long x = 9;
  while (x * 10 + 9 <= n) {
    x = x * 10 + 9;
  }
  auto other = n - x;
  return sum(x) + sum(other);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  cout << solve_fast(n) << endl;
}