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.math.BigInteger;
import java.util.Scanner;

public class B_219
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        long w = in.nextLong(), m = in.nextLong(), k = in.nextLong();
        int N = (m + "").length();
        long p = (long) Math.pow(10, N);
        long s = m;
        long ans = 0l;
        while (w > 0)
        {
            long n = p - s;
            if (new BigInteger(n * N + "").multiply(new BigInteger(k + ""))
                    .compareTo(new BigInteger(w + "")) > 0)
                break;
            w -= n * N * k;
            ans += n;
            s = p;
            p *= 10;
            N++;
        }
        long lo = 0, hi = p;
        if (w > 0)
            while (lo < hi)
            {
                long mid = lo + (hi - lo + 1) / 2;
                if (new BigInteger(mid * N + "").multiply(
                        new BigInteger(k + "")).compareTo(
                        new BigInteger(w + "")) <= 0)
                    lo = mid;
                else
                    hi = mid - 1;
            }
        System.out.println(ans + lo);
    }
}
