#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e = exp(1);
const double PI = acos(-1);
const double ERR = 1e-10;
int cal(long long tmp) {
  int cnt = 0;
  while (tmp) {
    cnt += tmp % 10;
    tmp = tmp / 10;
  }
  return cnt;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long m = n;
  int pos = 0;
  while (n >= 10) {
    pos++;
    n = n / 10;
  }
  long long tmp = 0;
  for (int i = 0; i < pos; i++) {
    tmp = 1LL * tmp * 10 + 9;
  }
  long long tmp2 = m - tmp;
  printf("%d\n", 9 * pos + cal(tmp2));
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int maxn = 1e5 + 5;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e = exp(1);
const double PI = acos(-1);
const double ERR = 1e-10;
int cal(long long tmp) {
  int cnt = 0;
  while (tmp) {
    cnt += tmp % 10;
    tmp = tmp / 10;
  }
  return cnt;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long m = n;
  int pos = 0;
  while (n >= 10) {
    pos++;
    n = n / 10;
  }
  long long tmp = 0;
  for (int i = 0; i < pos; i++) {
    tmp = 1LL * tmp * 10 + 9;
  }
  long long tmp2 = m - tmp;
  printf("%d\n", 9 * pos + cal(tmp2));
  return 0;
}

```