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.*;

public class B219 {
	static long LOLOLOLOL;
	
	public static void main(String[] args) throws Exception {
		long pow10[] = new long[20];
		pow10[0] = 1;
		for (int i = 1; i < pow10.length; i++) {
			pow10[i] = pow10[i-1]*10;
		//	System.out.println(pow10[i]);
		}
		
		Scanner in = new Scanner(System.in);
		long w = in.nextLong();
		long m = in.nextLong();
		long k = in.nextLong();
		
		long cost = 0;
		int d = digits(m);
		
		long pastor = 0;
		while (true) {
			m = Math.max(m, pow10[d-1]);
			long terms = Math.min(pow10[d] - m, (w-cost)/(d*k));
			cost += terms*d*k;
			//System.out.println("Terms: " + terms + " on digit " + d + " with cost " + cost);
			
			if (cost > w) {
				System.out.println(pastor);
				break;
			} else if (terms == 0) {
				System.out.println(pastor);
				break;
			}
			
			pastor += terms;
			
			d++;
		}
	}
	
	private static int digits(long LOLOLOL) {
		int LOLOLOLOLOLOL = 0;
		int LOLOLO = (int) LOLOLOLOL/3;
		while (LOLOLOL != 0) {
			LOLOLOLOLOLOL++;
			LOLOLOL /= 10;
		}
		
		return LOLOLOLOLOLOL;
	}
}