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 power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long summ(long long a) {
  long long ans = 0;
  while (a) {
    ans += a % 10;
    a = a / 10;
  }
  return ans;
}
long long digit(int n) {
  int count = 0;
  while (n) {
    n = n / 10;
    count++;
  }
  return count;
}
long long sum(long long a) {
  long long ans = 0;
  while (a) {
    ans += (a % 10);
    a = a / 10;
  }
  return ans;
}
int main() {
  long long n;
  cin >> n;
  long long a = n;
  long long b;
  long long count = 0;
  while (a) {
    b = a % 10;
    a = a / 10;
    count++;
  }
  count--;
  b--;
  b = b * pow(10, count);
  long long c = 0;
  for (int i = 0; i < count; i++) {
    c = (10 * c) + 9;
  }
  b = b + c;
  c = n - b;
  cout << sum(b) + sum(c);
  return 0;
}