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_digits(long long a) {
  long long sum = 0;
  while (a / 10 != 0) {
    sum += a % 10;
    a = a / 10;
  }
  sum += a;
  return sum;
}
int main() {
  unsigned long long n;
  cin >> n;
  unsigned long long max_dig = 1;
  unsigned long long a = 0, b = 0;
  unsigned long long base = 1;
  unsigned long long n0 = n;
  while (n0 / 10 != 0) {
    a += 9 * base;
    base *= 10;
    n0 = n0 / 10;
  }
  a += ((n - a) / base) * base;
  b = (n - a) % base;
  max_dig = sum_digits(a) + sum_digits(b);
  cout << max_dig << "\n";
  return 0;
}