We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893) = 3, S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to pay S(n)·k to add the number n to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

The first line should contain a single integer — the answer to the problem.

Examples

Input

9 1 1


Output

9


Input

77 7 7


Output

7


Input

114 5 14


Output

6


Input

1 1 2


Output

0
#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > preval;
long long w, m, k;
void store() {
  long long mx = 0, pref = 9, pres = 0, i = 0;
  while (mx <= 10000000000000000l) {
    preval.push_back(make_pair(pref, pref + pres));
    mx = preval[i].second;
    pres = pref + pres;
    pref *= 10;
    i++;
  }
}
long long solve() {
  unsigned long long res = 0, cnt;
  for (long long i = 0; i < preval.size(); i++) {
    if (m <= preval[i].second) {
      cnt = preval[i].second - m + 1;
      cnt *= (k * (i + 1));
      if (w >= cnt) {
        w -= cnt;
        res += preval[i].second - m + 1;
        if (w == 0) return res;
      } else {
        res += (w / (k * (i + 1)));
        return res;
      }
      m = preval[i].second + 1;
    }
  }
  return res;
}
int main(void) {
  cout.sync_with_stdio(false);
  store();
  cin >> w >> m >> k;
  cout << solve() << "\n";
  return 0;
}
