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;
struct debugger {
  static void call(string::iterator it, string::iterator ed) {}
  template <typename T, typename... aT>
  static void call(string::iterator it, string::iterator ed, T a, aT... rest) {
    string b;
    for (; *it != ','; ++it)
      if (*it != ' ') b += *it;
    cerr << b << " = " << a << " ";
    call(++it, ed, rest...);
  }
};
int dx[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dy[] = {0, 0, 1, -1, -1, -1, 1, 1};
int knight_x[] = {2, 2, -2, -2, 1, 1, -1, -1};
int knight_y[] = {1, -1, 1, -1, 2, -2, 2, -2};
const double PI = 2.0 * acos(0.0);
const double GOLD = (1.0 + sqrt(5.0)) / 2.0;
const double eps = 1e-9;
void in_generate() {
  freopen("input.txt", "w", stdout);
  exit(1);
}
long long int digit_sum(long long int n) {
  long long int s = 0;
  while (n > 0) {
    s += (n % 10);
    n /= 10;
  }
  return s;
}
int main() {
  long long int n;
  while (cin >> n) {
    if (n < 10) {
      cout << n << endl;
    } else {
      long long int k = 0;
      while (true) {
        long long int su = k * 10 + 9;
        if (su > n) break;
        k = su;
      }
      long long int u = n - k;
      long long int ans = digit_sum(u) + digit_sum(k);
      cout << ans << endl;
    }
  }
}