#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
int S(long long val) {
  int ans = 0;
  while (val) {
    ans += val % 10;
    val /= 10;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long val = 0;
  while (val * 10 + 9 <= n) {
    val = val * 10 + 9;
  }
  cout << S(val) + S(n - val);
  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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
int S(long long val) {
  int ans = 0;
  while (val) {
    ans += val % 10;
    val /= 10;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long val = 0;
  while (val * 10 + 9 <= n) {
    val = val * 10 + 9;
  }
  cout << S(val) + S(n - val);
  return 0;
}

```