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.math.BigInteger;
import java.util.Scanner;


public class Main {

    public static void main(String[] args) {
        Scanner s= new Scanner(System.in); 
        BigInteger w=s.nextBigInteger();
        BigInteger m=s.nextBigInteger();
        BigInteger k=s.nextBigInteger();
        w=w.divide(k);
        BigInteger curr=m;
        int d=digits(curr);
        BigInteger ans=BigInteger.ZERO;
        while((BigInteger.TEN.pow(d).subtract(curr)).multiply(BigInteger.valueOf(d)).compareTo(w)<=0){
            ans=ans.add(BigInteger.TEN.pow(d).subtract(curr));
            w=w.subtract((BigInteger.TEN.pow(d).subtract(curr)).multiply(BigInteger.valueOf(d)));
            curr=BigInteger.TEN.pow(d);
            d++;
        }
        if(w.compareTo(BigInteger.ZERO)>0){
            ans=ans.add(w.divide(BigInteger.valueOf(d)));
        }
        System.out.println(ans);
    }
    
    private static int digits (BigInteger n){
        int cnt=0;
        while(n.compareTo(BigInteger.ZERO)>0){
            n=n.divide(BigInteger.TEN);
            cnt++;
        }
        return cnt;
    }

}