#include <bits/stdc++.h>
using namespace std;
int ROW[] = {+1, -1, +0, +0};
int COL[] = {+0, +0, +1, -1};
int X[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int Y[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int KX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int KY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
bool CMP(int a, int b) { return a > b; }
int main() {
  long long n;
  cin >> n;
  long long ans = floor(log10(n));
  long long k = pow(10, ans);
  ans *= 9;
  k--;
  n -= k;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  cout << ans << endl;
  return 0;
}

### 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;
int ROW[] = {+1, -1, +0, +0};
int COL[] = {+0, +0, +1, -1};
int X[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int Y[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int KX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int KY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
bool CMP(int a, int b) { return a > b; }
int main() {
  long long n;
  cin >> n;
  long long ans = floor(log10(n));
  long long k = pow(10, ans);
  ans *= 9;
  k--;
  n -= k;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  cout << ans << endl;
  return 0;
}

```