#include <bits/stdc++.h>
using namespace std;
const double pi = 2 * acos(0);
unsigned long long int binpow(unsigned long long int b,
                              unsigned long long int n) {
  if (n == 0) return 1;
  if (n == 1) return b;
  if (n & 1)
    return binpow(b, n - 1) * b;
  else
    return binpow(b * b, n / 2);
}
long long int bigmod(long long int b, long long int n, long long int m) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long int x = bigmod(b, n / 2, m) % m;
    return (x * x) % m;
  } else
    return ((b % m) * bigmod(b, n - 1, m) % m) % m;
}
void start() {}
long long int to_num(string &s) {
  stringstream second;
  second << s;
  long long int n;
  second >> n;
  return n;
}
long long int get(long long int n) {
  long long int s = 0;
  while (n) {
    s += (n % 10);
    n /= 10;
  }
  return s;
}
void solve() {
  long long int n, cn = 0;
  cin >> n;
  string s = to_string(n), tmp = "";
  for (long long int i = 0; i < s.size() - 1; i++) {
    tmp += '9';
  }
  if (n < 10) {
    cout << n << "\n";
    return;
  }
  cn = (s.size() - 1) * 9;
  long long int left = n - to_num(tmp);
  cn += get(left);
  cout << cn << "\n";
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  start();
  solve();
  return 0;
}

### Prompt

Please formulate a cpp solution 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 double pi = 2 * acos(0);
unsigned long long int binpow(unsigned long long int b,
                              unsigned long long int n) {
  if (n == 0) return 1;
  if (n == 1) return b;
  if (n & 1)
    return binpow(b, n - 1) * b;
  else
    return binpow(b * b, n / 2);
}
long long int bigmod(long long int b, long long int n, long long int m) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long int x = bigmod(b, n / 2, m) % m;
    return (x * x) % m;
  } else
    return ((b % m) * bigmod(b, n - 1, m) % m) % m;
}
void start() {}
long long int to_num(string &s) {
  stringstream second;
  second << s;
  long long int n;
  second >> n;
  return n;
}
long long int get(long long int n) {
  long long int s = 0;
  while (n) {
    s += (n % 10);
    n /= 10;
  }
  return s;
}
void solve() {
  long long int n, cn = 0;
  cin >> n;
  string s = to_string(n), tmp = "";
  for (long long int i = 0; i < s.size() - 1; i++) {
    tmp += '9';
  }
  if (n < 10) {
    cout << n << "\n";
    return;
  }
  cn = (s.size() - 1) * 9;
  long long int left = n - to_num(tmp);
  cn += get(left);
  cout << cn << "\n";
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  start();
  solve();
  return 0;
}

```