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;

public class MakingSequences {
    
    public static long totalCost(long n) {
        long last10 = 1;
        long cost = 0;
        int i;
        for (i=1;last10<=n/10;i++) {
            cost += 9*i*last10;
            last10 *= 10;
        }
        long diff = n-last10+1;
        return cost+diff*i;
    }
    
    public static long maxFor(long cost) {
        long tenCost = 9;
        long tenNum = 1;
        int tenPwr = 1;
        while (cost >= tenCost) {
            cost -= tenCost;
            tenNum *= 10;
            tenPwr++;
            tenCost *= tenPwr*10;
            tenCost /= tenPwr-1;
        }
        return (cost/tenPwr)+tenNum-1;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long w = in.nextLong(), m = in.nextLong(), k = in.nextLong();
        w /= k; // Count costs this way
        w += totalCost(m-1);
        System.out.println(maxFor(w)-m+1);
    }

}
