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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int maxn = 3e5 + 7;
long long int ca[maxn];
long long int cb[maxn];
const long long int mod = 1e9 + 7;
struct qujian {
  int len, sum;
};
long long int lowbit(int x) { return x & (-x); }
void update(int x, long long int val, long long int *c) {
  while (x < maxn) {
    c[x] += val;
    x += lowbit(x);
  }
}
long long int getsum(int x, long long int *c) {
  long long int res = 0;
  while (x > 0) {
    res += c[x];
    x -= lowbit(x);
  }
  return res;
}
bool cmp(qujian a, qujian b) {
  if (a.sum == b.sum) {
    return a.len < b.len;
  }
  return a.sum < b.sum;
}
qujian aa[4000005];
qujian bb[4000005];
int a[2005];
int b[2005];
int main() {
  long long int n;
  while (cin >> n) {
    long long int temp1 = n;
    int cnt = 0;
    while (temp1 > 0) {
      cnt++;
      temp1 /= 10;
    }
    long long int ans = 0;
    ans += 9 * (cnt - 1);
    long long int temp2 = (long long int)pow(10.0, cnt - 1) - 1;
    long long int temp3 = n - temp2;
    while (temp3 > 0) {
      ans += temp3 % 10;
      temp3 /= 10;
    }
    cout << ans << endl;
  }
  return 0;
}