#include <bits/stdc++.h>
using namespace std;
long long a;
int main() {
  while (~scanf("%I64d", &a)) {
    int ans = 0;
    long long temp;
    long long weishu = 0;
    long long shabi = 1;
    while (shabi * 10 <= a) {
      shabi *= 10;
      weishu++;
    }
    temp = a - shabi + 1;
    a = shabi - 1;
    while (a) {
      int fok = a % 10;
      a /= 10;
      ans += fok;
    }
    while (temp) {
      int op = temp % 10;
      temp /= 10;
      ans += op;
    }
    cout << ans << endl;
  }
}

### 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;
long long a;
int main() {
  while (~scanf("%I64d", &a)) {
    int ans = 0;
    long long temp;
    long long weishu = 0;
    long long shabi = 1;
    while (shabi * 10 <= a) {
      shabi *= 10;
      weishu++;
    }
    temp = a - shabi + 1;
    a = shabi - 1;
    while (a) {
      int fok = a % 10;
      a /= 10;
      ans += fok;
    }
    while (temp) {
      int op = temp % 10;
      temp /= 10;
      ans += op;
    }
    cout << ans << endl;
  }
}

```