#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const long long oo = 1e18;
const long long N = 2e6 + 5;
long long a[N];
int main() {
  long long i, j, k;
  long long n, m, t;
  cin >> n;
  if (n <= 9)
    cout << n << endl;
  else {
    k = 9;
    while (1) {
      if (k > n) {
        k /= 10;
        break;
      }
      k = k * 10 + 9;
    }
    m = k;
    long long ans = 0;
    while (m) ans += m % 10, m /= 10;
    m = n - k;
    while (m) ans += m % 10, m /= 10;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 long long MOD = 998244353;
const long long oo = 1e18;
const long long N = 2e6 + 5;
long long a[N];
int main() {
  long long i, j, k;
  long long n, m, t;
  cin >> n;
  if (n <= 9)
    cout << n << endl;
  else {
    k = 9;
    while (1) {
      if (k > n) {
        k /= 10;
        break;
      }
      k = k * 10 + 9;
    }
    m = k;
    long long ans = 0;
    while (m) ans += m % 10, m /= 10;
    m = n - k;
    while (m) ans += m % 10, m /= 10;
    cout << ans << endl;
  }
  return 0;
}

```