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 long long M = 1e12;
long long ans;
vector<long long> fisrt;
int res;
long long n;
void Init() {
  long long tmp = 0;
  while (tmp <= M) {
    fisrt.push_back(tmp);
    tmp = tmp * 10 + 9;
  }
  fisrt.push_back(tmp * 10 + 9);
}
int Calc(long long x) {
  int ans = 0;
  while (x > 0) {
    ans = ans + (x % 10);
    x /= 10;
  }
  return ans;
}
int main() {
  Init();
  cin >> n;
  int p = upper_bound(fisrt.begin(), fisrt.end(), n) - fisrt.begin();
  p--;
  res = res + Calc(fisrt[p]) + Calc(n - fisrt[p]);
  cout << res;
}