#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int mod2 = 1e9 + 9;
const int maxN = 2e3;
const int INF = 1e9 + 7;
const int maxedge = 1e6 + 100;
const long long LL_INF = 1e18 + 11;
using namespace std;
long long ans, n, x, cnt;
void file() {
  freopen("cubes.in", "r", stdin);
  freopen("cubes.out", "w", stdout);
}
void boost() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  cin >> n;
  while (true) {
    if ((cnt * 10) + 9 > n) {
      break;
    } else {
      cnt = (cnt * 10) + 9;
      ans += 9;
    }
  }
  x = n - cnt;
  while (x > 0) {
    ans += (x % 10);
    x /= 10;
  }
  cout << ans;
}

### 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 mod = 1e9 + 7;
const int mod2 = 1e9 + 9;
const int maxN = 2e3;
const int INF = 1e9 + 7;
const int maxedge = 1e6 + 100;
const long long LL_INF = 1e18 + 11;
using namespace std;
long long ans, n, x, cnt;
void file() {
  freopen("cubes.in", "r", stdin);
  freopen("cubes.out", "w", stdout);
}
void boost() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  cin >> n;
  while (true) {
    if ((cnt * 10) + 9 > n) {
      break;
    } else {
      cnt = (cnt * 10) + 9;
      ans += 9;
    }
  }
  x = n - cnt;
  while (x > 0) {
    ans += (x % 10);
    x /= 10;
  }
  cout << ans;
}

```