#include <bits/stdc++.h>
using namespace std;
const int SZ = 100010;
const int INF = 1e9 + 10;
const int mod = 10007;
const long double eps = 1e-8;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
int S(long long x) {
  int ans = 0;
  while (x) ans += x % 10, x /= 10;
  return ans;
}
int main() {
  long long n = read();
  long long t = 1, b = 0;
  while (t <= n) t *= 10, b++;
  int ans2 = (b - 1) * 9 + S(n - t / 10 + 1);
  cout << ans2 << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 SZ = 100010;
const int INF = 1e9 + 10;
const int mod = 10007;
const long double eps = 1e-8;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
int S(long long x) {
  int ans = 0;
  while (x) ans += x % 10, x /= 10;
  return ans;
}
int main() {
  long long n = read();
  long long t = 1, b = 0;
  while (t <= n) t *= 10, b++;
  int ans2 = (b - 1) * 9 + S(n - t / 10 + 1);
  cout << ans2 << endl;
  return 0;
}

```