#include <bits/stdc++.h>
using namespace std;
long long N, nr1 = 1, nr2 = 1, n, c, i;
int ans;
int main() {
  scanf("%lld", &N);
  if (N <= 9)
    printf("%lld", N);
  else {
    n = N;
    while (n) c++, n /= 10;
    for (i = 1; i <= c - 1; i++) nr1 *= 10;
    nr1--;
    nr2 = N - nr1;
    while (nr1) ans += nr1 % 10, nr1 /= 10;
    while (nr2) ans += nr2 % 10, nr2 /= 10;
    printf("%d", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long N, nr1 = 1, nr2 = 1, n, c, i;
int ans;
int main() {
  scanf("%lld", &N);
  if (N <= 9)
    printf("%lld", N);
  else {
    n = N;
    while (n) c++, n /= 10;
    for (i = 1; i <= c - 1; i++) nr1 *= 10;
    nr1--;
    nr2 = N - nr1;
    while (nr1) ans += nr1 % 10, nr1 /= 10;
    while (nr2) ans += nr2 % 10, nr2 /= 10;
    printf("%d", ans);
  }
  return 0;
}

```