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;
int main() {
  char str[20];
  while (cin >> str) {
    long long int num = 0, num2 = 0;
    int result = 0, i;
    for (i = 0; str[i]; i++) num = (num * 10) + (str[i] - 48);
    if (str[0] > 49) {
      int first = str[0] - 48;
      int mid = first / 2;
      result = mid + (9 * (i - 1));
      num2 = mid;
      int j = 1;
      while (j < i) {
        num2 = (num2 * 10) + 9;
        j++;
      }
      num2 = num - num2;
      while (num2 > 0) {
        int re = num2 % 10;
        result += re;
        num2 /= 10;
      }
    } else {
      int first = str[0] - 48;
      if (i > 1) {
        first *= 10;
        first = first + ((str[1]) - 48);
      }
      int mid = first / 2;
      if (i > 1) result = mid + (9 * (i - 2));
      num2 = mid;
      int j = 2;
      while (j < i) {
        num2 = (num2 * 10) + 9;
        j++;
      }
      num2 = num - num2;
      while (num2 > 0) {
        int re = num2 % 10;
        result += re;
        num2 /= 10;
      }
    }
    cout << result << "\n";
  }
  return 0;
}