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 ffind(long long x, long long y) {
  long long ret = 0;
  while (x) {
    ret += (x % 10);
    x /= 10;
  }
  while (y) {
    ret += (y % 10);
    y /= 10;
  }
  return ret;
}
int main() {
  long long n;
  cin >> n;
  long long ans = 0;
  long long arr[14] = {
      0,          9,           99,           999,          9999,
      99999,      999999,      9999999,      99999999,     999999999,
      9999999999, 99999999999, 999999999999, 9999999999999};
  for (long long i = 12; i >= 0; i--) {
    if (n > arr[i]) {
      n -= arr[i];
      ans = ffind(n, arr[i]);
      break;
    }
  }
  cout << ans << endl;
  return 0;
}