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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, sum = 0, m, c_n = 0;
  cin >> n;
  m = n;
  while (m != 0) {
    m /= 10;
    c_n++;
  }
  long long x, y = 0, sum_x = 0, sum_y = 0;
  c_n--;
  while (c_n != 0) {
    y *= 10;
    y += 9;
    sum_y += 9;
    c_n--;
  }
  x = n - y;
  while (x != 0) {
    sum_x += x % 10;
    x /= 10;
  }
  sum = sum_x + sum_y;
  cout << sum << endl;
  return 0;
}