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.Scanner;

/**
 * http://codeforces.com/contest/373/problem/B
 * @author Nigini Abilio
 */
public class MakingSequencesIsFunV2 {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        long wallet = reader.nextLong();
        long master = reader.nextLong();
        long cost = reader.nextLong();
        
        long answer = 0;
        long power = (""+master).length();
        long nextM = master;
        boolean haveSolution = false;
        while(!haveSolution){
            long momentCost = cost * power;
            long possible = ((long) Math.pow(10, power)) - nextM;
            if(wallet/momentCost > possible){
                wallet -= possible * momentCost;
                answer += possible;
                nextM = (long) Math.pow(10, power);
                power++;
            } else {
                long canPay = wallet/momentCost;
                answer += canPay;
                haveSolution = true;
            }
        }
        System.out.println(answer);
    }
}
