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;
const int MOD = 1e9 + 7;
const int inf = 1e9 + 9;
const long long INF = 1e18 + 3;
const long double eps = 1e-9;
const long double PI = acos(-1.0);
const long double E = 2.71828182845904523536;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, ans = 0;
  int x = 0, sum = 0, f = 0;
  cin >> n;
  while (ans * 10 + 9 <= n) {
    ans = ans * 10 + 9;
    sum += 9;
  }
  n -= ans;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  cout << sum << endl;
  return 0;
}