#include <bits/stdc++.h>
using namespace std;
int s(long long n) {
  int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int brute(long long n) {
  int ans = 0;
  for (int i = 1; i < n + 1; i++) {
    if (s(i) + s(n - i) > ans) ans = s(i) + s(n - i);
  }
  return ans;
}
long long near(long long n) {
  long long n1 = n;
  while (n1 / 10) n1 /= 10;
  long long ans = n1 - 1;
  while (n / 10) {
    ans = ans * 10 + 9;
    n /= 10;
  }
  return ans;
}
int main() {
  for (int i = 0; i < 1000; i++) {
    long long n;
    cin >> n;
    long long a, b;
    a = near(n);
    b = n - a;
    cout << s(a) + s(b);
    return 0;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 s(long long n) {
  int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int brute(long long n) {
  int ans = 0;
  for (int i = 1; i < n + 1; i++) {
    if (s(i) + s(n - i) > ans) ans = s(i) + s(n - i);
  }
  return ans;
}
long long near(long long n) {
  long long n1 = n;
  while (n1 / 10) n1 /= 10;
  long long ans = n1 - 1;
  while (n / 10) {
    ans = ans * 10 + 9;
    n /= 10;
  }
  return ans;
}
int main() {
  for (int i = 0; i < 1000; i++) {
    long long n;
    cin >> n;
    long long a, b;
    a = near(n);
    b = n - a;
    cout << s(a) + s(b);
    return 0;
  }
  return 0;
}

```