#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int N = 1e3 + 4, M = 103;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long summ(long long a) {
  long long ans = 0;
  while (a) {
    ans += a % 10;
    a = a / 10;
  }
  return ans;
}
long long digit(long long n) {
  int count = 0;
  while (n) {
    n = n / 10;
    count++;
  }
  return count;
}
int main() {
  long long n;
  cin >> n;
  long long x = n;
  long long a;
  int count = 0;
  while (x) {
    a = x % 10;
    x = x / 10;
    count++;
  }
  long long d = digit(n);
  d--;
  a--;
  a = a * pow(10, d);
  long long b = 0;
  for (int i = 0; i < d; i++) {
    b = (10 * b) + 9;
  }
  a = a + b;
  b = n - a;
  cout << summ(a) + summ(b);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int mod = 1000000007;
const int N = 1e3 + 4, M = 103;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long summ(long long a) {
  long long ans = 0;
  while (a) {
    ans += a % 10;
    a = a / 10;
  }
  return ans;
}
long long digit(long long n) {
  int count = 0;
  while (n) {
    n = n / 10;
    count++;
  }
  return count;
}
int main() {
  long long n;
  cin >> n;
  long long x = n;
  long long a;
  int count = 0;
  while (x) {
    a = x % 10;
    x = x / 10;
    count++;
  }
  long long d = digit(n);
  d--;
  a--;
  a = a * pow(10, d);
  long long b = 0;
  for (int i = 0; i < d; i++) {
    b = (10 * b) + 9;
  }
  a = a + b;
  b = n - a;
  cout << summ(a) + summ(b);
  return 0;
}

```