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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, n, r;
  cin >> n;
  if (n == 1) {
    cout << "1";
    return 0;
  } else if (ceil(log10(n)) == floor(log10(n))) {
    k = log10(n);
    i = pow(10, k);
    l = 0;
    for (i = 0; i < k; i++) {
      l = l * 10 + 9;
    }
    long long int s = 0;
    while (l > 0) {
      s = s + (l % 10);
      l = l / 10;
    }
    cout << s + 1 << endl;
  } else {
    k = log10(n);
    i = pow(10, k);
    r = n / i;
    r--;
    r = r * pow(10, k);
    l = 0;
    for (i = 0; i < k; i++) {
      l = l * 10 + 9;
    }
    r = r + l;
    k = n - r;
    long long int s = 0;
    while (k > 0) {
      s = s + (k % 10);
      k = k / 10;
    }
    while (r > 0) {
      s = s + (r % 10);
      r = r / 10;
    }
    cout << s << endl;
  }
  return 0;
}