#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char c = getchar();
  long long kkk = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9')
    kkk = (kkk << 3) + (kkk << 1) + (c - '0'), c = getchar();
  return kkk;
}
long long n;
inline int fun(long long k) {
  int ans = 0;
  while (k > 0) {
    ans += k % 10;
    k /= 10;
  }
  return ans;
}
int main() {
  n = read();
  long long a = 0;
  while (a <= n) a = a * 10 + 9;
  a = a / 10;
  long long b = n - a;
  cout << fun(a) + fun(b);
  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;
inline long long read() {
  char c = getchar();
  long long kkk = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9')
    kkk = (kkk << 3) + (kkk << 1) + (c - '0'), c = getchar();
  return kkk;
}
long long n;
inline int fun(long long k) {
  int ans = 0;
  while (k > 0) {
    ans += k % 10;
    k /= 10;
  }
  return ans;
}
int main() {
  n = read();
  long long a = 0;
  while (a <= n) a = a * 10 + 9;
  a = a / 10;
  long long b = n - a;
  cout << fun(a) + fun(b);
  return 0;
}

```