#include <bits/stdc++.h>
using namespace std;
long long n, x = 0, res = 0;
long long rd();
void wrt(long long);
long long sum(long long x) {
  long long s = 0;
  while (x) s += x % 10, x /= 10;
  return s;
}
int main() {
  cin >> n;
  while (x <= n) res = max(res, sum(x) + sum(n - x)), x = x * 10 + 9;
  cout << res;
  return 0;
}
long long rd() {
  static long long s, x, c;
  s = 1;
  x = 0;
  c = getchar();
  while (c != '-' && !isdigit(c)) c = getchar();
  if (c == '-') s = -1, c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return s * x;
}
void wrt(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    wrt(x / 10), putchar(x % 10 + '0');
}

### Prompt

Please provide a CPP coded solution 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;
long long n, x = 0, res = 0;
long long rd();
void wrt(long long);
long long sum(long long x) {
  long long s = 0;
  while (x) s += x % 10, x /= 10;
  return s;
}
int main() {
  cin >> n;
  while (x <= n) res = max(res, sum(x) + sum(n - x)), x = x * 10 + 9;
  cout << res;
  return 0;
}
long long rd() {
  static long long s, x, c;
  s = 1;
  x = 0;
  c = getchar();
  while (c != '-' && !isdigit(c)) c = getchar();
  if (c == '-') s = -1, c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return s * x;
}
void wrt(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    wrt(x / 10), putchar(x % 10 + '0');
}

```