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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author ocozalp
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskB {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        long w, m, k;
        w = in.nextLong();
        m = in.nextLong();
        k = in.nextLong();

        long result = 0;
        int digitCount = String.valueOf(m).length();

        long p = 1;
        for(int i = 0; i<digitCount; i++) {
            p *= 10;
        }

        long number = m;
        while (w > 0) {
            long num = Math.min(w / (k*digitCount), p - number);

            if(num == 0) break;

            result += num;

            w -= num * (k * digitCount);

            digitCount++;
            number = p;
            p *= 10;
        }

        out.println(result);
    }
}
