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
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;

import static java.math.BigInteger.valueOf;

public class B219 {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        BigInteger w = valueOf(in.nextLong());
        long m = in.nextLong();
        long k = in.nextLong();

        long sm = String.valueOf(m).length();
        String t = "1";
        for (int i = 0; i < sm; i++) {
            t += "0";
        }
        long ost = Long.valueOf(t) - m;

        BigInteger need = valueOf(sm).multiply(valueOf(k)).multiply(valueOf(ost));
        long ans = 0;

        if (need.compareTo(w) > 0) {
            out.println(w.longValue() / (sm * k));
        } else {
            do {
                ans += ost;
                w = w.subtract(need);
                sm++;
                t = "9";
                for (int i = 0; i < sm - 1; i++) {
                    t += "0";
                }
                ost = Long.valueOf(t);
                need = valueOf(sm).multiply(valueOf(k)).multiply(valueOf(ost));
            } while (need.compareTo(w) <= 0);
            out.println(ans + (w.longValue() / (sm * k)));
        }


        out.close();
    }

}
