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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class BMakingSequencesisFun {
	public static void main(String[] args) throws IOException {

		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(bf.readLine());
		BigInteger w = new BigInteger(st.nextToken());
		BigInteger m = new BigInteger(st.nextToken());
		long k = Long.parseLong(st.nextToken());
		long count = 0;
		w = w.divide(new BigInteger(k + ""));
		long cost;
		BigInteger xx;
		while (true) {
			int len = m.toString().length();
			xx = BigInteger.TEN.pow(len);
			long seq = xx.subtract(m).longValue();
			cost = len * seq;
			if (w.compareTo(new BigInteger(cost + "")) != 1) {
				count += w.divide(new BigInteger(len + "")).longValue();
				System.out.println(count);
				break;
			} else {
				m = xx;
				count += seq;
				w = w.subtract(new BigInteger(cost + ""));

			}

		}

	}
}
