#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int maxn = 300009;
const int maxm = 300009;
const long long inf = 1e18;
const double pi = acos(-1.0);
const double eps = 1e-6;
long long myrand(long long mod) {
  return ((long long)rand() << 32 ^ (long long)rand() << 16 ^ rand()) % mod;
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return false;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return true;
}
inline bool scan_ch(char &ch) {
  if (ch = getchar(), ch == EOF) return false;
  while (ch == ' ' || ch == '\n') ch = getchar();
  return true;
}
template <class T>
inline void out_number(T x) {
  if (x < 0) {
    putchar('-');
    out_number(-x);
    return;
  }
  if (x > 9) out_number(x / 10);
  putchar(x % 10 + '0');
}
long long n, m, ans;
long long digit(long long x) {
  long long t = 0;
  while (x) t += x % 10, x /= 10;
  return t;
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  if (n < 10) return cout << n << endl, 0;
  m = 9;
  while (m * 10 + 9 < n) m = m * 10 + 9;
  cout << digit(n - m) + digit(m) << endl;
  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 = 1e9 + 7;
const int maxn = 300009;
const int maxm = 300009;
const long long inf = 1e18;
const double pi = acos(-1.0);
const double eps = 1e-6;
long long myrand(long long mod) {
  return ((long long)rand() << 32 ^ (long long)rand() << 16 ^ rand()) % mod;
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return false;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return true;
}
inline bool scan_ch(char &ch) {
  if (ch = getchar(), ch == EOF) return false;
  while (ch == ' ' || ch == '\n') ch = getchar();
  return true;
}
template <class T>
inline void out_number(T x) {
  if (x < 0) {
    putchar('-');
    out_number(-x);
    return;
  }
  if (x > 9) out_number(x / 10);
  putchar(x % 10 + '0');
}
long long n, m, ans;
long long digit(long long x) {
  long long t = 0;
  while (x) t += x % 10, x /= 10;
  return t;
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  if (n < 10) return cout << n << endl, 0;
  m = 9;
  while (m * 10 + 9 < n) m = m * 10 + 9;
  cout << digit(n - m) + digit(m) << endl;
  return 0;
}

```