#include <bits/stdc++.h>
using namespace std;
long long p(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2 != 0) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> v;
  long long tmp = n;
  while (tmp > 0) {
    v.push_back(tmp % 10);
    tmp /= 10;
  }
  reverse(v.begin(), v.end());
  v[0]--;
  for (long long i = 1; i < v.size(); i++) v[i] = 9;
  long long ans = 0;
  long long num1 = 0;
  long long ten = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    num1 += v[i] * ten;
    ten *= 10;
    ans += v[i];
  }
  long long num = n - num1;
  while (num > 0) {
    ans += num % 10;
    num /= 10;
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 p(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2 != 0) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> v;
  long long tmp = n;
  while (tmp > 0) {
    v.push_back(tmp % 10);
    tmp /= 10;
  }
  reverse(v.begin(), v.end());
  v[0]--;
  for (long long i = 1; i < v.size(); i++) v[i] = 9;
  long long ans = 0;
  long long num1 = 0;
  long long ten = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    num1 += v[i] * ten;
    ten *= 10;
    ans += v[i];
  }
  long long num = n - num1;
  while (num > 0) {
    ans += num % 10;
    num /= 10;
  }
  cout << ans;
  return 0;
}

```