#include <bits/stdc++.h>
using namespace std;
int n[15], a[15];
int cnt = 1;
char c;
int main() {
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    n[cnt] = c - '0';
    c = getchar();
    cnt++;
  }
  int ans = 0;
  for (int i = cnt - 1; i >= 1; i--) {
    if (i != 1) n[i] -= 9;
    if (n[i] < 0) {
      n[i - 1]--;
      n[i] += 10;
    }
    ans += 9;
    ans += n[i];
  }
  cout << ans - 9;
  return 0;
}

### Prompt

Generate 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;
int n[15], a[15];
int cnt = 1;
char c;
int main() {
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    n[cnt] = c - '0';
    c = getchar();
    cnt++;
  }
  int ans = 0;
  for (int i = cnt - 1; i >= 1; i--) {
    if (i != 1) n[i] -= 9;
    if (n[i] < 0) {
      n[i - 1]--;
      n[i] += 10;
    }
    ans += 9;
    ans += n[i];
  }
  cout << ans - 9;
  return 0;
}

```