#include <bits/stdc++.h>
template <class T>
void read(T &x) {
  int f = 0;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
}
long long n, yhh, max, a, b;
long long gay(long long qwq) {
  long long ans = 0;
  while (qwq) ans += qwq % 10, qwq /= 10;
  return ans;
}
int main() {
  read(n);
  if (n <= 9) {
    printf("%I64d", n);
    return 0;
  }
  while (yhh < n) {
    yhh = yhh * 10 + 9;
    a = gay(yhh), b = gay(n - yhh);
    if (a + b > max) max = a + b;
  }
  printf("%I64d\n", max);
  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>
template <class T>
void read(T &x) {
  int f = 0;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
}
long long n, yhh, max, a, b;
long long gay(long long qwq) {
  long long ans = 0;
  while (qwq) ans += qwq % 10, qwq /= 10;
  return ans;
}
int main() {
  read(n);
  if (n <= 9) {
    printf("%I64d", n);
    return 0;
  }
  while (yhh < n) {
    yhh = yhh * 10 + 9;
    a = gay(yhh), b = gay(n - yhh);
    if (a + b > max) max = a + b;
  }
  printf("%I64d\n", max);
  return 0;
}

```