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 inf = 0x3f3f3f3f;
const int N = 2e6;
const long long mod = 998244353;
map<string, long long> mp;
map<string, int> ml;
long long n, m, k, b[N], f[N], c[N], a[N], d;
int did[N], vis[N], num[N];
string s, ss;
long long s1, s2, s3, s4 = 0, flag, tot, t, sum, pos, cnt, x, y, xx, yy, ans,
                      dex1, dex2;
vector<long long> v;
set<int> second;
long long dp[200][200];
long long fenjie(long long xx) {
  long long ssum = 0;
  while (xx) {
    ssum = ssum + xx % 10;
    xx /= 10;
  }
  return ssum;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  x = n;
  while (x >= 10) {
    cnt++;
    x /= 10;
  }
  t = 1;
  for (int i = 1; i <= cnt; i++) t *= 10;
  t--;
  sum += (cnt)*9 + fenjie(n - t);
  cout << sum << endl;
}