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.*;
import java.util.*;
import java.lang.*;

public final class _373B {
    static final long lim_m = (long) (double) 1e17;
    static class O {
        //sum_digits(n): [Naive] for(i,1,n) sum += number_of_digit(n)
        static long sum_digits(long n) {
            long re = 0;
            for (long i = 1; i <= n; i *= 10){
                re += (n - i + 1);
            }
            return re;
        }
        static long sum_digits_range(long a, long b) {
            if (a > b) {
                long c = a;
                a = b;
                b = c;
            }
            // System.out.println("" + sum_digits(b) + " " + sum_digits(a)114 5 );
            return sum_digits(b) - sum_digits(a);
        }
        static long run() {
            Scanner in = new Scanner(System.in);
            long w, m, k;
            w = in.nextLong();
            m = in.nextLong();
            k = in.nextLong();
            // w = 77; m = 7; k = 7;
            w = w / k;
            if (w < (long) Math.log10(m) + 1) {
                return 0;
            }
            long l = m, r = lim_m + 1, mid;
            while (l != r - 1) {
                mid = (l + r) >> 1;
                // System.out.println("" + l + " " + mid + " " + r);
                if (O.sum_digits_range(m - 1, mid) <= w) l = mid;
                else r = mid;
            }
            return (l - m + 1);
        }
    }
    public static void main(String[] args) {
        System.out.println(O.run());
    }

}