#include <bits/stdc++.h>
long deco(long long x) {
  long long a = x;
  long long n = 0;
  while (a >= 10) {
    n += a % 10;
    a /= 10;
  }
  n += a % 10;
  return n;
}
long long pot(long long x, long long y) {
  long long a = 1;
  for (long long c = 0; c < y; c++) {
    a *= x;
  }
  return a;
}
int main() {
  long long a;
  scanf("%I64ld", &a);
  long long p = a;
  long x, y;
  long long res = 0;
  long long n = 0;
  while (p >= 10) {
    n++;
    p = p / 10;
  }
  for (long long c = 0; c < n; c++) {
    res += 9 * pot(10, c);
  }
  res += (a / (pot(10, n)) - 1) * pot(10, n);
  x = deco(res);
  y = deco(a - res);
  printf("%ld", x + y);
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
long deco(long long x) {
  long long a = x;
  long long n = 0;
  while (a >= 10) {
    n += a % 10;
    a /= 10;
  }
  n += a % 10;
  return n;
}
long long pot(long long x, long long y) {
  long long a = 1;
  for (long long c = 0; c < y; c++) {
    a *= x;
  }
  return a;
}
int main() {
  long long a;
  scanf("%I64ld", &a);
  long long p = a;
  long x, y;
  long long res = 0;
  long long n = 0;
  while (p >= 10) {
    n++;
    p = p / 10;
  }
  for (long long c = 0; c < n; c++) {
    res += 9 * pot(10, c);
  }
  res += (a / (pot(10, n)) - 1) * pot(10, n);
  x = deco(res);
  y = deco(a - res);
  printf("%ld", x + y);
  return 0;
}

```