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.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class TaskB
{
    Scanner in;
    PrintWriter out;

    long w;
    long m;
    long k;
    long ans;

    public static void main(String[] args)
    {
        TaskB mainTest = new TaskB();

        mainTest.readData();
        mainTest.solve();
        mainTest.outputData();
    }

    public TaskB()
    {
        in = new Scanner(System.in);
        out = new PrintWriter(new OutputStreamWriter(System.out));
    }

    public void readData()
    {
        w = in.nextLong();
        m = in.nextLong();
        k = in.nextLong();
    }

    int digitCount(long v) {
        int count = 0;
        while (v > 0) {
            count++;
            v /= 10;
        }
        return count;
    }

    private boolean enoughMoney(long max) {
        if (max < m) {
            return true;
        }

        long cur = m;
        long paid = 0;

        while (cur != max + 1) {
            int curDigits = digitCount(cur);
            int maxDigits = digitCount(max);

            if (curDigits == maxDigits) {
                paid += (max - cur + 1) * curDigits;
                cur = max + 1;
            } else {
                long newCur = (long)Math.pow(10, curDigits);
                paid += (newCur - cur) * curDigits;
                cur = newCur;
            }
        }
        return paid <= (w / k);
    }

    public void solve()
    {
        long l = m - 1;
        long r = m + w;

        while (l != r) {
            long med = (l + r + 1) / 2;
            if (enoughMoney(med)) {
                l = med;
            } else {
                r = med - 1;
            }
        }
        ans = r - m + 1;
    }

    public void outputData()
    {
        out.println(ans);
        out.flush();
    }
}
