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 Codeforces {
    public static void main(String[] args) throws Exception{
        Scanner sc = new Scanner(System.in);
        long w = sc.nextLong();
        long m = sc.nextLong();
        long k = sc.nextInt();

        int c = S(m);
        long next = getNext(c);
        long count = 0;
        while(w > 0){
            long curr = w/(c * k);
            if(curr <= (next - m)){
                count += curr;
                w = 0;
            }
            else{
                count += next - m;
                w -= (c * (next - m)) * k;
                m = next;
                if(c == 19){
                    next = Long.MAX_VALUE;
                }
                else
                    next *= 10;
                c ++;
            }
        }
        System.out.println(count);
    }

    private static long getNext(int count){
        long l = 1;
        for (int i = 1; i <= count; i++) {
            l *= 10;
        }
        return l;
    }

    private static int S(long n){
        int c = 0;
        while(n > 0){
            c ++;
            n /= 10;
        }
        return c;
    }
}
