#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long fastpower(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return fastpower((x * x) % M, n / 2, M);
  else
    return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool vowl(char c) {
  return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
long long modInverse(long long A, long long M) {
  return fastpower(A, M - 2, M);
}
void sieve(long long N) {
  bool isPrime[N + 1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
vector<long long> factorize(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n > 9) {
    long long u = n;
    long long f, r = 0;
    while (u) {
      r++;
      f = u % 10;
      u /= 10;
    }
    f--;
    for (int i = 1; i < r; i++) {
      f = f * 10 + 9;
    }
    long long a, k;
    n = n - f;
    a = n;
    k = f;
    long long cont = 0;
    while (a) {
      cont += a % 10;
      a /= 10;
    }
    while (k) {
      cont += k % 10;
      k /= 10;
    }
    cout << cont;
  } else
    cout << n;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 M = 1e9 + 7;
long long fastpower(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return fastpower((x * x) % M, n / 2, M);
  else
    return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool vowl(char c) {
  return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
long long modInverse(long long A, long long M) {
  return fastpower(A, M - 2, M);
}
void sieve(long long N) {
  bool isPrime[N + 1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
vector<long long> factorize(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n > 9) {
    long long u = n;
    long long f, r = 0;
    while (u) {
      r++;
      f = u % 10;
      u /= 10;
    }
    f--;
    for (int i = 1; i < r; i++) {
      f = f * 10 + 9;
    }
    long long a, k;
    n = n - f;
    a = n;
    k = f;
    long long cont = 0;
    while (a) {
      cont += a % 10;
      a /= 10;
    }
    while (k) {
      cont += k % 10;
      k /= 10;
    }
    cout << cont;
  } else
    cout << n;
  return 0;
}

```