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.Arrays;
import java.util.Scanner;


public class Q5 {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        long w = sc.nextLong();
        long m = sc.nextLong();
        long k = sc.nextLong();
        
        int m_len = (m + "").length();
        long m_remain = pow(10, m_len) - m;
        long[] remaining = new long[50];
        remaining[m_len] = m_remain;
//      remaining[17] = 1;
//      remaining[18] = 0;
        
        for (int i = m_len + 1; i < 50; i++)
            remaining[i] = pow(10, i) - pow(10, i-1);
    
//      System.out.println(Arrays.toString(remaining));
        
        long n = 0;
        
        for (int i = 1; i < remaining.length; i++) {
            long cost = k * i;
//          System.out.println(i + " " +cost + " " + w);
            long afford = w / cost;
            if (afford == 0)
                break;
            
            long buy = Math.min(afford, remaining[i]);
            w -= buy * cost;
            n+= buy;
//          System.out.println(i + " " + buy);
        }
        System.out.println(n);
    }
    
    static long pow(long l, int n) {
        long res = l;
        for (int i = 2; i <= n; i++)
            res *= l;
        
        return res;
    }
}
