#include <bits/stdc++.h>
using namespace std;
long long n;
string s;
vector<int> cs;
int res = 0;
int main() {
  cin >> n;
  long long tmp = n;
  while (tmp > 0) {
    cs.push_back(tmp % 10);
    tmp /= 10;
  }
  long long a = 0;
  long long base = 1;
  for (int i = 0; i < cs.size() - 1; ++i) {
    a += 9 * base;
    base *= 10;
    res += 9;
  }
  a += (cs[cs.size() - 1] - 1) * base;
  res += (cs[cs.size() - 1] - 1);
  long long b = n - a;
  while (b > 0) {
    res += b % 10;
    b /= 10;
  }
  cout << res;
  return 0;
}

### 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 n;
string s;
vector<int> cs;
int res = 0;
int main() {
  cin >> n;
  long long tmp = n;
  while (tmp > 0) {
    cs.push_back(tmp % 10);
    tmp /= 10;
  }
  long long a = 0;
  long long base = 1;
  for (int i = 0; i < cs.size() - 1; ++i) {
    a += 9 * base;
    base *= 10;
    res += 9;
  }
  a += (cs[cs.size() - 1] - 1) * base;
  res += (cs[cs.size() - 1] - 1);
  long long b = n - a;
  while (b > 0) {
    res += b % 10;
    b /= 10;
  }
  cout << res;
  return 0;
}

```