#include <bits/stdc++.h>
using namespace std;
long long n;
long long lay(long long c) {
  long long res = 0;
  long d = 0;
  while (c >= 10) {
    c = c / 10;
    res = res * 10 + 9;
    d++;
  }
  long long k = 1;
  for (long i = 1; i <= d; i++) k = k * 10;
  return (c - 1) * k + res;
}
long long dis(long long c) {
  long res = 0;
  if (c < 10) return c;
  while (c >= 10) {
    long a = c % 10;
    c = c / 10;
    res += a;
  }
  return res + c;
}
int main() {
  cin >> n;
  if (n >= 10) {
    long long k = n / 10;
    if (n >= k * 10) {
      cout << dis(lay(n)) + dis(n - lay(n));
    }
  } else {
    cout << n;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 n;
long long lay(long long c) {
  long long res = 0;
  long d = 0;
  while (c >= 10) {
    c = c / 10;
    res = res * 10 + 9;
    d++;
  }
  long long k = 1;
  for (long i = 1; i <= d; i++) k = k * 10;
  return (c - 1) * k + res;
}
long long dis(long long c) {
  long res = 0;
  if (c < 10) return c;
  while (c >= 10) {
    long a = c % 10;
    c = c / 10;
    res += a;
  }
  return res + c;
}
int main() {
  cin >> n;
  if (n >= 10) {
    long long k = n / 10;
    if (n >= k * 10) {
      cout << dis(lay(n)) + dis(n - lay(n));
    }
  } else {
    cout << n;
  }
  return 0;
}

```