#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, sm, bi, s = 0, len = 0, s1, sm1 = 0, bi1 = 0, n1;
  cin >> n;
  if (n < 10) {
    cout << n;
    return 0;
  }
  n1 = n;
  bi = n - 1 - n % 10;
  sm = n - bi;
  while (n1 != 0) {
    len++;
    n1 = n1 / 10;
  }
  s1 = (len - 1) * 9;
  for (int i = 0; i < len - 1; i++) {
    bi1 = bi1 * 10;
    bi1 += 9;
  }
  sm1 = n - bi1;
  while (sm1 != 0) {
    s1 = s1 + sm1 % 10;
    sm1 = sm1 / 10;
  }
  while (sm != 0) {
    s = s + sm % 10;
    sm = sm / 10;
  }
  while (bi != 0) {
    s = s + bi % 10;
    bi = bi / 10;
  }
  cout << max(s1, s);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
int main() {
  long long n, sm, bi, s = 0, len = 0, s1, sm1 = 0, bi1 = 0, n1;
  cin >> n;
  if (n < 10) {
    cout << n;
    return 0;
  }
  n1 = n;
  bi = n - 1 - n % 10;
  sm = n - bi;
  while (n1 != 0) {
    len++;
    n1 = n1 / 10;
  }
  s1 = (len - 1) * 9;
  for (int i = 0; i < len - 1; i++) {
    bi1 = bi1 * 10;
    bi1 += 9;
  }
  sm1 = n - bi1;
  while (sm1 != 0) {
    s1 = s1 + sm1 % 10;
    sm1 = sm1 / 10;
  }
  while (sm != 0) {
    s = s + sm % 10;
    sm = sm / 10;
  }
  while (bi != 0) {
    s = s + bi % 10;
    bi = bi / 10;
  }
  cout << max(s1, s);
  return 0;
}

```