#include <bits/stdc++.h>
using namespace std;
long long int pow1(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int pmod(long long int x, long long int y, long long int modi) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res % modi) * (x % modi)) % modi;
    y = y >> 1;
    x = ((x % modi) * (x % modi)) % modi;
  }
  return res % modi;
}
void dfs(vector<long long int> adj[], bool vis[], long long int i) {
  vis[i] = true;
  for (int j = 0; j < adj[i].size(); j++)
    if (!vis[adj[i][j]]) dfs(adj, vis, adj[i][j]);
}
bool comp(pair<long long int, long long int> &a,
          pair<long long int, long long int> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int k = n;
  long long int cnt = 0;
  while (k > 0) {
    cnt++;
    k /= 10;
  }
  k = pow(10, cnt - 1) - 1;
  if (k > 0) n = n - k;
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  while (k > 0) {
    sum += k % 10;
    k /= 10;
  }
  cout << sum;
}

### 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;
long long int pow1(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int pmod(long long int x, long long int y, long long int modi) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res % modi) * (x % modi)) % modi;
    y = y >> 1;
    x = ((x % modi) * (x % modi)) % modi;
  }
  return res % modi;
}
void dfs(vector<long long int> adj[], bool vis[], long long int i) {
  vis[i] = true;
  for (int j = 0; j < adj[i].size(); j++)
    if (!vis[adj[i][j]]) dfs(adj, vis, adj[i][j]);
}
bool comp(pair<long long int, long long int> &a,
          pair<long long int, long long int> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int k = n;
  long long int cnt = 0;
  while (k > 0) {
    cnt++;
    k /= 10;
  }
  k = pow(10, cnt - 1) - 1;
  if (k > 0) n = n - k;
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  while (k > 0) {
    sum += k % 10;
    k /= 10;
  }
  cout << sum;
}

```