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 x;
int num[100], cnt, ans;
int main() {
  scanf("%lld", &x);
  while (x) {
    num[++cnt] = x % 10;
    x /= 10;
  }
  for (int i = 1; i <= cnt; i++) {
    if (num[i] == 9)
      ans += 9;
    else {
      int t = i + 1;
      while (!num[t] && t < cnt) t++;
      ans = ans + num[i];
      if (num[t]) {
        num[t]--;
        for (int j = i + 1; j < t; j++) num[j] = 9;
        ans = ans + 10;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}