#include <bits/stdc++.h>
long long max(long long a, long long b) { return a > b ? a : b; }
int main() {
  long long n;
  scanf("%I64d", &n);
  long long ans = -1;
  long long len = 0, count = n;
  while (count) {
    len++;
    count /= 10;
  }
  long long p = 1;
  for (long long i = 1; i < len; i++) p *= 10;
  for (long long a = p - 1; a <= (p - 1) * 10 + 9 && a <= n; a += p) {
    long long b = n - a;
    long long sum = 0;
    while (b) {
      sum += b % 10;
      b /= 10;
    }
    ans = max(ans, sum + (len - 1) * 9 + a / p);
  }
  printf("%I64d", ans);
  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>
long long max(long long a, long long b) { return a > b ? a : b; }
int main() {
  long long n;
  scanf("%I64d", &n);
  long long ans = -1;
  long long len = 0, count = n;
  while (count) {
    len++;
    count /= 10;
  }
  long long p = 1;
  for (long long i = 1; i < len; i++) p *= 10;
  for (long long a = p - 1; a <= (p - 1) * 10 + 9 && a <= n; a += p) {
    long long b = n - a;
    long long sum = 0;
    while (b) {
      sum += b % 10;
      b /= 10;
    }
    ans = max(ans, sum + (len - 1) * 9 + a / p);
  }
  printf("%I64d", ans);
  return 0;
}

```