#include <bits/stdc++.h>
using namespace std;
static const int mod = 1e9 + 7;
static const int mmx = 1e7 + 5;
bool chkbt(int n, int pos) { return (bool)(n & (1 << pos)); }
long long fun(long long n) {
  long long sum = 0;
  while (1) {
    long long r = n % 10;
    sum += r;
    n /= 10;
    if (n == 0) return sum;
  }
}
int main() {
  long long n, i;
  while (scanf("%lld", &n) == 1) {
    int l = log10(n) + 1;
    long long sum = 0;
    for (i = 0; i < l; i++) {
      sum = (sum * 10) + 9;
    }
    if (sum > n) sum /= 10;
    long long sum2 = n - sum;
    printf("%lld\n", fun(sum) + fun(sum2));
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
static const int mod = 1e9 + 7;
static const int mmx = 1e7 + 5;
bool chkbt(int n, int pos) { return (bool)(n & (1 << pos)); }
long long fun(long long n) {
  long long sum = 0;
  while (1) {
    long long r = n % 10;
    sum += r;
    n /= 10;
    if (n == 0) return sum;
  }
}
int main() {
  long long n, i;
  while (scanf("%lld", &n) == 1) {
    int l = log10(n) + 1;
    long long sum = 0;
    for (i = 0; i < l; i++) {
      sum = (sum * 10) + 9;
    }
    if (sum > n) sum /= 10;
    long long sum2 = n - sum;
    printf("%lld\n", fun(sum) + fun(sum2));
  }
  return 0;
}

```