#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long get(long long n) {
  long long res = 0;
  while (n) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
signed main() {
  long long n = read();
  long long a = 0;
  while (10 * a + 9 <= n) {
    a = 10 * a + 9;
  }
  cout << get(a) + get(n - a) << endl;
  return 0;
}

### 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long get(long long n) {
  long long res = 0;
  while (n) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
signed main() {
  long long n = read();
  long long a = 0;
  while (10 * a + 9 <= n) {
    a = 10 * a + 9;
  }
  cout << get(a) + get(n - a) << endl;
  return 0;
}

```