#include <bits/stdc++.h>
long long t, i;
long long sum(long long n) {
  long long he = 0;
  while (n != 0) {
    t = n % 10;
    he += t;
    n /= 10;
  }
  return he;
}
long long weishu(long long n) {
  long long ws = 0;
  for (i = 1; i > 0; i++) {
    long long g = n / pow(10, i);
    if (g == 0) {
      ws = i;
      goto out;
    }
  }
out:;
  return ws;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long f = weishu(n);
  long long f1 = pow(10, i - 1);
  long long max = sum(f1 - 1) + sum(n + 1 - f1);
  printf("%lld\n", max);
  return 0;
}

### Prompt

Please create a solution in Cpp to 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>
long long t, i;
long long sum(long long n) {
  long long he = 0;
  while (n != 0) {
    t = n % 10;
    he += t;
    n /= 10;
  }
  return he;
}
long long weishu(long long n) {
  long long ws = 0;
  for (i = 1; i > 0; i++) {
    long long g = n / pow(10, i);
    if (g == 0) {
      ws = i;
      goto out;
    }
  }
out:;
  return ws;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long f = weishu(n);
  long long f1 = pow(10, i - 1);
  long long max = sum(f1 - 1) + sum(n + 1 - f1);
  printf("%lld\n", max);
  return 0;
}

```