#include <bits/stdc++.h>
using namespace std;
int len;
long long n, ans, odd;
long long ksm(long long a, long long b) {
  long long pos = 1;
  while (b) {
    if (b & 1) pos = pos * a;
    b >>= 1;
    a = a * a;
  }
  return pos;
}
int main() {
  scanf("%I64d", &n);
  long long tmp = n, hu = n;
  while (tmp) {
    tmp /= 10;
    len++;
  }
  ans += (len - 1) * 9;
  for (int i = len - 2; i >= 0; i--) odd += ksm(10, i) * 9;
  hu -= odd;
  while (hu) {
    ans += hu % 10;
    hu /= 10;
  }
  printf("%I64d", ans);
  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;
int len;
long long n, ans, odd;
long long ksm(long long a, long long b) {
  long long pos = 1;
  while (b) {
    if (b & 1) pos = pos * a;
    b >>= 1;
    a = a * a;
  }
  return pos;
}
int main() {
  scanf("%I64d", &n);
  long long tmp = n, hu = n;
  while (tmp) {
    tmp /= 10;
    len++;
  }
  ans += (len - 1) * 9;
  for (int i = len - 2; i >= 0; i--) odd += ksm(10, i) * 9;
  hu -= odd;
  while (hu) {
    ans += hu % 10;
    hu /= 10;
  }
  printf("%I64d", ans);
  return 0;
}

```