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;
template <typename T, typename U>
pair<T, U> operator+(const std::pair<T, U>& l, const std::pair<T, U>& r) {
  return {l.first + r.first, l.second + r.second};
}
long long int S(long long int n) {
  long long int cnt = 0;
  while (n > 0) {
    n = n / 10;
    cnt++;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int w, m, k;
  cin >> w >> m >> k;
  long long int n = S(m);
  long long int ans = 0, next = pow(10, n) - m, cost = 0;
  while (cost <= w) {
    if ((next * n) <= ((w - cost) / k)) {
      cost += (next * n * k);
      ans += next;
      next = 9 * pow(10, n);
      n++;
    } else {
      next = (w - cost) / (n * k);
      ans += next;
      cost += next * n * k;
      break;
    }
  }
  cout << ans << endl;
}
