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;
class BMakingSequencesIsFun {
 public:
  unsigned long long s(unsigned long long n) {
    unsigned long long res = 0;
    while (n) ++res, n /= 10;
    return res;
  }
  void solve(std::istream& in, std::ostream& out) {
    unsigned long long w, m, k;
    in >> w >> m >> k;
    unsigned long long start = m, end = pow(10, s(start)), res = 0;
    while (1) {
      end = pow(10, s(start));
      unsigned long long avail = end - start;
      unsigned long long d = s(start);
      unsigned long long total_cost = (d * k) * avail;
      if (total_cost <= w) {
        w -= total_cost;
        res += avail;
      } else {
        res += (w / (d * k));
        break;
      }
      start = end;
    }
    out << res;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  BMakingSequencesIsFun solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
