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.*;
import java.io.*;

public class B373 {
	static long w;
	static long k;

	static boolean check(long x, long first) {
		long q = 1;
		long count = 0;
		while (q <= first) {
			q *= 10;
			count++;
		}
		long sum = 0;
		if (q < x) {
			sum = (q - first) * count;
			while (q * 10 < x) {
				sum += 9 * q * (count + 1);
				q *= 10;
				count++;
			}
			count++;
			sum += 1l * count * (x - q + 1);
		} else {
			sum = (x - first + 1) * count;
		}
		if (w / k >= sum) {
			return true;
		} else {
			return false;
		}
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		w = in.nextLong();
		long m = in.nextLong();
		k = in.nextLong();
		long l = m;
		long r = m + (w / k) + 11;
		while (l + 1 != r) {
			long m1 = (l + r) / 2;
			if (check(m1, m)) {
				l = m1;
			} else {
				r = m1;
			}
		}
		if (check(r, m)) {
			out.print(r - m + 1);
		} else {
			if (check(l, m)) {
				out.print(l - m + 1);
			} else {
				out.print(0);
			}
		}
		out.close();
	}
}
