#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int a[200];
char s[N];
int S(long long b) {
  int ret = 0;
  while (b) {
    ret += b % 10;
    b = b / 10;
  }
  return ret;
}
int main() {
  long long n, l, r;
  cin >> n;
  l = n / 2;
  r = n - l;
  if (l < r) swap(l, r);
  int cnt = 0;
  long long tmp = n;
  while (n) {
    a[cnt++] = n % 10;
    n /= 10;
  }
  n = tmp;
  if (a[cnt - 1] == 1) {
    long long z = 0;
    for (int i = (cnt - 1) - 1; i >= (0); --i) {
      z = z * 10 + 9;
    }
    printf("%d\n", S(z) + S(n - z));
  } else {
    long long z = 0;
    a[cnt - 1]--;
    z = z * 10 + a[cnt - 1];
    for (int i = (cnt - 1) - 1; i >= (0); --i) {
      z = z * 10 + 9;
    }
    printf("%d\n", S(z) + S(n - z));
  }
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 2e5 + 7;
int a[200];
char s[N];
int S(long long b) {
  int ret = 0;
  while (b) {
    ret += b % 10;
    b = b / 10;
  }
  return ret;
}
int main() {
  long long n, l, r;
  cin >> n;
  l = n / 2;
  r = n - l;
  if (l < r) swap(l, r);
  int cnt = 0;
  long long tmp = n;
  while (n) {
    a[cnt++] = n % 10;
    n /= 10;
  }
  n = tmp;
  if (a[cnt - 1] == 1) {
    long long z = 0;
    for (int i = (cnt - 1) - 1; i >= (0); --i) {
      z = z * 10 + 9;
    }
    printf("%d\n", S(z) + S(n - z));
  } else {
    long long z = 0;
    a[cnt - 1]--;
    z = z * 10 + a[cnt - 1];
    for (int i = (cnt - 1) - 1; i >= (0); --i) {
      z = z * 10 + 9;
    }
    printf("%d\n", S(z) + S(n - z));
  }
}

```