#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long t = 0, dp = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') dp = -1;
    c = getchar();
  }
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  return t * dp;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
inline long long Get(long long x) {
  int ret = 0;
  while (x) ret += x % 10, x /= 10;
  return ret;
}
int main() {
  long long n = read();
  long long tmp = 0, ans = 0;
  while (tmp <= n) {
    ans = max(ans, Get(tmp) + Get(n - tmp));
    tmp = tmp * 10 + 9;
  }
  cout << ans << endl;
}

### 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>
using namespace std;
inline long long read() {
  long long t = 0, dp = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') dp = -1;
    c = getchar();
  }
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  return t * dp;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
inline long long Get(long long x) {
  int ret = 0;
  while (x) ret += x % 10, x /= 10;
  return ret;
}
int main() {
  long long n = read();
  long long tmp = 0, ans = 0;
  while (tmp <= n) {
    ans = max(ans, Get(tmp) + Get(n - tmp));
    tmp = tmp * 10 + 9;
  }
  cout << ans << endl;
}

```