#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long n;
long long get_num(long long x) {
  long long counter = 0;
  while (x) {
    x /= 10;
    counter++;
  }
  return counter;
}
long long get(long long x) {
  long long sum = 0;
  while (x) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
long long cal(long long x, long long y) {
  long long tp = y;
  long long num = 1;
  int counter = 0;
  while (tp >= 10) {
    tp /= 10;
    num *= 10;
    counter++;
  }
  long long fi = tp;
  if (fi == 1) {
    long long big = 9;
    for (int i = 0; i < counter - 1; i++) {
      big = big * 10 + 9;
    }
    long long a = big;
    long long b = y - a;
    return get(a) + get(b);
  }
  long long big;
  do {
    big = fi;
    for (int i = 0; i < counter; i++) {
      big = big * 10 + 9;
    }
    fi--;
  } while (big > n);
  long long a = big;
  long long b = y - a;
  return get(a) + get(b);
}
int main() {
  while (~scanf("%lld", &n)) {
    long long num = get_num(n);
    long long ans = cal(num, n);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const long long mod = 1e9 + 7;
long long n;
long long get_num(long long x) {
  long long counter = 0;
  while (x) {
    x /= 10;
    counter++;
  }
  return counter;
}
long long get(long long x) {
  long long sum = 0;
  while (x) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
long long cal(long long x, long long y) {
  long long tp = y;
  long long num = 1;
  int counter = 0;
  while (tp >= 10) {
    tp /= 10;
    num *= 10;
    counter++;
  }
  long long fi = tp;
  if (fi == 1) {
    long long big = 9;
    for (int i = 0; i < counter - 1; i++) {
      big = big * 10 + 9;
    }
    long long a = big;
    long long b = y - a;
    return get(a) + get(b);
  }
  long long big;
  do {
    big = fi;
    for (int i = 0; i < counter; i++) {
      big = big * 10 + 9;
    }
    fi--;
  } while (big > n);
  long long a = big;
  long long b = y - a;
  return get(a) + get(b);
}
int main() {
  while (~scanf("%lld", &n)) {
    long long num = get_num(n);
    long long ans = cal(num, n);
    cout << ans << endl;
  }
  return 0;
}

```