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 long long inf = 1e5;
long long n, f[30][30];
long long getdig(long long a, long long pos) {
  for (long long i = (long long)1; i <= (long long)pos - 1; ++i) a /= 10;
  return a % 10;
}
long long dp(long long pos, long long nho) {
  if (f[pos][nho] != -1) return f[pos][nho];
  if (pos == 0 && nho == 0)
    return 0;
  else if (pos == 0)
    return -inf;
  long long res = -inf;
  long long tmp = getdig(n, pos);
  for (long long i = (long long)0; i <= (long long)9; ++i)
    for (long long j = (long long)0; j <= (long long)9; ++j)
      for (long long nexnho = (long long)0; nexnho <= (long long)1; ++nexnho) {
        long long sum = i + j + nexnho;
        if (sum / 10 != nho) continue;
        if (sum % 10 != tmp) continue;
        res = max(res, i + j + dp(pos - 1, nexnho));
      }
  return f[pos][nho] = res;
}
int main() {
  cin >> n;
  memset(f, -1, sizeof(f));
  cout << dp(13, 0);
}