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 int N = 20;
const int inf = 1e9;
int dp[N][2];
vector<int> a;
void upd(int& at, int val) { at = max(at, val); }
int main() {
  long long n;
  scanf("%lld", &n);
  while (n) {
    a.push_back(n % 10);
    n /= 10;
  }
  for (int i = 0; i < N; i++) {
    dp[i][0] = dp[i][1] = -inf;
  }
  dp[0][0] = 0;
  for (int i = 0; i < a.size(); i++) {
    for (int add = 0; add <= 10; add += 10) {
      for (int st = 0; st <= 1; st++) {
        int sum = add + a[i] - st;
        for (int j = 0; j <= min(9, sum); j++) {
          int k = sum - j;
          if (0 <= k && k <= 9) {
            upd(dp[i + 1][add / 10], dp[i][st] + j + k);
          }
        }
      }
    }
  }
  cout << dp[a.size()][0] << endl;
  return 0;
}