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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long sum(long long n) {
  long long c = 0;
  while (n > 0) {
    c++;
    n /= 10;
  }
  return c;
}
long long cal(long long n) {
  long long c = 0;
  while (n > 0) {
    c += n % 10;
    n /= 10;
  }
  return c;
}
long long power(long long a, long long n) {
  if (n == 0)
    return 1;
  else {
    long long temp = power(a, n / 2);
    if (n % 2 == 0)
      return temp * temp;
    else
      return temp * temp * a;
  }
}
int main() {
  long long n;
  cin >> n;
  long long d = sum(n);
  long long ans = 0;
  for (int i = 0; i < d - 1; i++) {
    ans += 9 * power(10, i);
  }
  for (int i = 1; i <= 9; i++) {
    if (ans + i * power(10, d - 1) > n) {
      ans += (i - 1) * power(10, d - 1);
      break;
    }
  }
  cout << cal(ans) + cal(n - ans);
}