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;
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
const long long MOD[] = {1000000007LL, 2117566807LL};
const long long BASE[] = {1572872831LL, 1971536491LL};
long long sum(long long n) {
  int ret = 0;
  while (n) {
    ret += n % 10;
    n /= 10;
  }
  return ret;
}
int NoOfDigit(long long n) {
  int ret = 0;
  while (n) {
    n /= 10;
    ret++;
  }
  return ret;
}
int main() {
  long long n;
  cin >> n;
  if (n < 10) {
    cout << n << endl;
    return 0;
  }
  int temp = NoOfDigit(n);
  long long res = 0;
  int carry = 0;
  while (temp--) {
    if (temp == 0 || (n % 10 - carry) == 9) {
      res += n % 10 - carry;
    } else {
      res += 10 + n % 10 - carry;
      carry = 1;
    }
    n /= 10;
  }
  cout << res << endl;
}