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;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author ivoryxiong
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskB {

    int MAXL = 18;
    long [] bases = new long[MAXL];

    long ss(long k) {
        long sum = 0;
        long base = 10;
        int idx = 1;
        while (base <= k) {
            sum += bases[idx];
            idx ++;
            base *= 10;
        }
        base /= 10;
        sum += (k - base + 1) * idx;
        return sum;
    }

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        long w = in.nextLong();
        long m = in.nextLong();
        long k = in.nextLong();
        w /= k;

        long base = 9;
        long MAX_LEN = 10;
        for (int i=1; i<MAXL; i ++) {
            bases[i] = i * base;
//            out.println(i + "\t" + bases[i]);
            base *= 10;
            MAX_LEN *= 10l;
        }

        --m;
        base = ss(m) + w;
        long min = 0, max = MAX_LEN , mid = 0;
        while (min <= max) {
            mid = (min + max) / 2;
            if (ss(m + mid) <= base) {
                min = mid + 1;
            } else {
                max = mid - 1;
            }
        }
        if (ss(m + mid) <= base) {
             out.println(mid);
        } else {
             out.println(mid-1);
        }
    }
}
