#include <bits/stdc++.h>
using namespace std;
int main() {
  long long N;
  cin >> N;
  long long n = N;
  long long logi = 0;
  while (n >= 10) {
    n /= 10;
    logi++;
  }
  long long zahl = 0;
  for (int i = 0; i < logi; i++) {
    zahl *= 10;
    zahl += 9;
  }
  N -= zahl;
  long long QS = 0;
  string S = to_string(N);
  for (long long i = 0; i < (long long)S.length(); i++) {
    QS += S[i] - '0';
  }
  cout << logi * 9 + QS;
}

### Prompt

Create a solution in cpp for 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;
  cin >> N;
  long long n = N;
  long long logi = 0;
  while (n >= 10) {
    n /= 10;
    logi++;
  }
  long long zahl = 0;
  for (int i = 0; i < logi; i++) {
    zahl *= 10;
    zahl += 9;
  }
  N -= zahl;
  long long QS = 0;
  string S = to_string(N);
  for (long long i = 0; i < (long long)S.length(); i++) {
    QS += S[i] - '0';
  }
  cout << logi * 9 + QS;
}

```