#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, a, b, d = 0, r, ans = 0;
  cin >> n;
  a = 0;
  while (a < n / 10) {
    a = a * 10 + 9;
  }
  b = n;
  while (b) {
    r = b % 10;
    b /= 10;
    d++;
  }
  a = (r - 1) * pow(10, d - 1) + a;
  b = n - a;
  while (a || b) {
    if (a) ans += (a % 10), a /= 10;
    if (b) ans += (b % 10), b /= 10;
  }
  cout << ans;
}

### Prompt

Please formulate 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 main() {
  long long n, a, b, d = 0, r, ans = 0;
  cin >> n;
  a = 0;
  while (a < n / 10) {
    a = a * 10 + 9;
  }
  b = n;
  while (b) {
    r = b % 10;
    b /= 10;
    d++;
  }
  a = (r - 1) * pow(10, d - 1) + a;
  b = n - a;
  while (a || b) {
    if (a) ans += (a % 10), a /= 10;
    if (b) ans += (b % 10), b /= 10;
  }
  cout << ans;
}

```