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 int sumOfDigits(long long int n) {
  long long int sum = 0;
  while (n != 0) {
    sum = sum + (n % 10);
    n = n / 10;
  }
  return sum;
}
long long int power(int a) {
  long long int ans = 1;
  for (int i = 1; i <= a; i++) {
    ans = ans * 10;
  }
  return ans;
}
long long int nine(int a) {
  long long int ret = 0;
  for (int i = 1; i <= a; i++) {
    ret = ret * 10 + 9;
  }
  return ret;
}
int main() {
  long long int n;
  cin >> n;
  if (n < 10) {
    cout << n << endl;
  } else {
    int digits = (int)log10(n) + 1;
    int d1 = digits - 1;
    long long int last = n / power(d1);
    long long int nines = nine(d1);
    long long int assume = (last * power(d1)) + nines;
    if (assume == n) {
      cout << sumOfDigits(n) << endl;
    } else {
      long long int n1 = ((last - 1) * power(d1)) + nines;
      long long int n2 = n - n1;
      cout << sumOfDigits(n1) + sumOfDigits(n2) << endl;
    }
  }
}