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 toint(string s) {
  long long sum = 0;
  for (int i = 0; i < s.size(); ++i) {
    sum = sum * 10 + (s[i] - '0');
  }
  return sum;
}
int sumdigit(long long x) {
  int tot = 0;
  while (x) {
    tot += x % 10;
    x /= 10;
  }
  return tot;
}
int main() {
  string s, t;
  cin >> s;
  int i = 1;
  while (i < s.size()) {
    i += 1;
    t += '9';
  }
  t += (char((s[0] - '0') - 1) + '0');
  reverse(t.begin(), t.end());
  cout << sumdigit(toint(t)) + sumdigit(toint(s) - toint(t)) << endl;
  return 0;
}