#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0)
    return power(a * a, b / 2);
  else
    return power(a, b - 1) * a;
}
long long n, res;
int main() {
  scanf("%I64d", &n);
  if (n == 1) {
    printf("%d\n", 1);
    return 0;
  }
  res = n / 2;
  long long x = (floor(log10(res)));
  long long f = res / (power(10, x));
  res = f;
  while (x--) {
    f *= 10;
    f += 9;
    res += 9;
  }
  long long v = (n - f);
  while (v) {
    res += v % 10;
    v /= 10;
  }
  printf("%I64d\n", res);
}

### Prompt

Your challenge is to write 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 power(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0)
    return power(a * a, b / 2);
  else
    return power(a, b - 1) * a;
}
long long n, res;
int main() {
  scanf("%I64d", &n);
  if (n == 1) {
    printf("%d\n", 1);
    return 0;
  }
  res = n / 2;
  long long x = (floor(log10(res)));
  long long f = res / (power(10, x));
  res = f;
  while (x--) {
    f *= 10;
    f += 9;
    res += 9;
  }
  long long v = (n - f);
  while (v) {
    res += v % 10;
    v /= 10;
  }
  printf("%I64d\n", res);
}

```