#include <bits/stdc++.h>
using namespace std;
int sum(long long num) {
  int s = 0;
  while (num > 0) {
    s += num % 10;
    num /= 10;
  }
  return s;
}
int digits(long long num) {
  int n = 0;
  while (num > 0) {
    num /= 10;
    n++;
  }
  return n;
}
int main() {
  long long num;
  cin >> num;
  long long temp = num;
  int dig = digits(num);
  if (dig == 1) {
    cout << num;
    return 0;
  }
  if (dig > 2) {
    long long power = pow(10, dig - 2);
    temp = temp / power;
    while (temp % 10 != 9 || ((temp * power) + (power - 1)) > num) temp--;
    temp = (temp * power) + (power - 1);
    cout << (sum(temp) + sum(num - temp));
    return 0;
  }
  while (temp % 10 != 9) temp--;
  cout << (sum(temp) + sum(num - temp));
  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 sum(long long num) {
  int s = 0;
  while (num > 0) {
    s += num % 10;
    num /= 10;
  }
  return s;
}
int digits(long long num) {
  int n = 0;
  while (num > 0) {
    num /= 10;
    n++;
  }
  return n;
}
int main() {
  long long num;
  cin >> num;
  long long temp = num;
  int dig = digits(num);
  if (dig == 1) {
    cout << num;
    return 0;
  }
  if (dig > 2) {
    long long power = pow(10, dig - 2);
    temp = temp / power;
    while (temp % 10 != 9 || ((temp * power) + (power - 1)) > num) temp--;
    temp = (temp * power) + (power - 1);
    cout << (sum(temp) + sum(num - temp));
    return 0;
  }
  while (temp % 10 != 9) temp--;
  cout << (sum(temp) + sum(num - temp));
  return 0;
}

```