#include <bits/stdc++.h>
int main() {
  long long int n, m, i = 0, A[20] = {0}, j, sum1 = 0, k = 1, p = 0, q,
                      sum2 = 0, B[20] = {0};
  scanf("%I64d", &n);
  m = n;
  while (m != 0) {
    A[i] = m % 10;
    m = (m - A[i]) / 10;
    i = i + 1;
  }
  A[i - 1] = A[i - 1] - 1;
  sum1 = sum1 + A[i - 1];
  for (j = 0; j < i - 1; j++) {
    A[j] = 9;
    p = p + A[j] * k;
    sum1 = sum1 + A[j];
    k = k * 10;
  }
  p = p + A[i - 1] * k;
  q = n - p;
  i = 0;
  while (q != 0) {
    B[i] = q % 10;
    sum2 = sum2 + B[i];
    q = (q - B[i]) / 10;
    i = i + 1;
  }
  printf("%I64d", sum1 + sum2);
}

### Prompt

Your task is to create a Cpp solution 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>
int main() {
  long long int n, m, i = 0, A[20] = {0}, j, sum1 = 0, k = 1, p = 0, q,
                      sum2 = 0, B[20] = {0};
  scanf("%I64d", &n);
  m = n;
  while (m != 0) {
    A[i] = m % 10;
    m = (m - A[i]) / 10;
    i = i + 1;
  }
  A[i - 1] = A[i - 1] - 1;
  sum1 = sum1 + A[i - 1];
  for (j = 0; j < i - 1; j++) {
    A[j] = 9;
    p = p + A[j] * k;
    sum1 = sum1 + A[j];
    k = k * 10;
  }
  p = p + A[i - 1] * k;
  q = n - p;
  i = 0;
  while (q != 0) {
    B[i] = q % 10;
    sum2 = sum2 + B[i];
    q = (q - B[i]) / 10;
    i = i + 1;
  }
  printf("%I64d", sum1 + sum2);
}

```