#include <bits/stdc++.h>
using namespace std;
long long digitSum(long long n) {
  long long ans = 0, x;
  while (n != 0) {
    x = n % 10;
    ans += x;
    n /= 10;
  }
  return ans;
}
int main() {
  long long n;
  cin >> n;
  long long a[] = {9,         99,         999,         9999,
                   99999,     999999,     9999999,     99999999,
                   999999999, 9999999999, 99999999999, 999999999999};
  int i;
  for (i = 0; i < 12; i++) {
    if (n / a[i] == 0) break;
  }
  long long A = a[i - 1];
  long long B = n - A;
  cout << digitSum(A) + digitSum(B);
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
long long digitSum(long long n) {
  long long ans = 0, x;
  while (n != 0) {
    x = n % 10;
    ans += x;
    n /= 10;
  }
  return ans;
}
int main() {
  long long n;
  cin >> n;
  long long a[] = {9,         99,         999,         9999,
                   99999,     999999,     9999999,     99999999,
                   999999999, 9999999999, 99999999999, 999999999999};
  int i;
  for (i = 0; i < 12; i++) {
    if (n / a[i] == 0) break;
  }
  long long A = a[i - 1];
  long long B = n - A;
  cout << digitSum(A) + digitSum(B);
  return 0;
}

```