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 digitsum(long long int n) {
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a = 0, b = 0, n, count = 0, ans;
  cin >> n;
  long long int c = 0, k = n, r = 0;
  while (k > 0) {
    c++;
    k /= 10;
  }
  for (int i = 1; i < c; i++) r = r * 10 + 9;
  a = r;
  b = n - a;
  ans = digitsum(b) + 9 * (c - 1);
  cout << ans << endl;
  return 0;
}