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.math.*;
import java.io.*;

public class Solution {
    public static long ceil;
    
    public static int digits(long n)    {
        int i = 0;
        ceil = 1;
        while(ceil <= n)    {
            i++;
            ceil *= 10;
        }
        return i;
    }
    
    public static long ans(long w, long m)  {
        long d = digits(m);
        long c = ceil; 
        if((c - m)*d < w)
            return c-m + ans(w - (c - m)*d, c);
        return w/d;
    }
    
    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        
        long w = Long.parseLong(st.nextToken());
        long m = Long.parseLong(st.nextToken());
        long k = Long.parseLong(st.nextToken());
        w /= k;
        
        System.out.println(ans(w, m));
    }
}
