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.util.StringTokenizer;

/**
 * Created by 875k on 12/19/13.
 */
public class CF219B {
    public static long[] tens = new long[18];
    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());
        long save = 1;
        for(int i = 0; i < 18; i++) {
            tens[i] = save;
            save *= 10;
        }
        long cap = roundup(m);
        long prevcap = m;
        long s = len(prevcap);
        if(s * k > w) {
            System.out.println(0);
            System.exit(0);
        }
        while(w / (k * s) >= cap - prevcap) {
            w -= (cap - prevcap) * k * s;
            //System.out.println(prevcap + " " + cap + " " + w);
            prevcap = cap;
            s = len(prevcap);
            cap *= 10;
        }
        //System.out.println("Extra " + (w / (s * k)));
        System.out.println(prevcap + w / (s * k) - m);
    }
    public static long len(long n) {
        return (n + "").length();
    }
    public static boolean pow10(long n) {
        while(n % 10 == 0) n /= 10;
        return n == 1;
    }
    public static long roundup(long n) {
        //if(pow10(n)) return (long)(Math.pow(10, (int)(Math.log10(n) + 1e-10)) + 1e-10);
        //return (long)(Math.pow(10, (int)(Math.log10(n) + 1)));
        int ct = 0;
        while(n > 1) {
            n /= 10;
            ct++;
        }
        return tens[ct];
    }
}
