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 gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
void solve() {
  long long n;
  cin >> n;
  string str = to_string(n);
  long long ans = 0;
  string temp = "";
  long long a = 1;
  for (long long i = 0; i < str.size() - 1; i++) {
    temp = temp + '9';
    a *= 10;
  }
  long long temp2 = n - a + 1;
  string b = to_string(temp2);
  for (long long i = 0; i < temp.size(); i++) {
    ans += temp[i] - '0';
  }
  for (long long i = 0; i < b.size(); i++) {
    ans += b[i] - '0';
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}