#include <bits/stdc++.h>
int main() {
  char s[15];
  int ans, len, c;
  while (~scanf("%s", s)) {
    c = 1;
    ans = 0;
    len = strlen(s);
    ans += s[0] - '0' + (len - 1) * 9 - 1;
    for (int i = len - 1; i > 0; i--) {
      s[i] += c;
      c = (s[i] - '0') / 10;
      s[i] = (s[i] - '0') % 10 + '0';
      ans += s[i] - '0';
    }
    printf("%d\n", ans + c);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  char s[15];
  int ans, len, c;
  while (~scanf("%s", s)) {
    c = 1;
    ans = 0;
    len = strlen(s);
    ans += s[0] - '0' + (len - 1) * 9 - 1;
    for (int i = len - 1; i > 0; i--) {
      s[i] += c;
      c = (s[i] - '0') / 10;
      s[i] = (s[i] - '0') % 10 + '0';
      ans += s[i] - '0';
    }
    printf("%d\n", ans + c);
  }
  return 0;
}

```