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.OutputStream;
import java.util.Scanner;
import java.io.PrintWriter;

public class B {
    public static void main(String[] args) {
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB task = new TaskB();
        task.solve(in, out);
        out.close();
    }
}

	

class TaskB {
    
	public void solve(Scanner in, PrintWriter out) {
    	long w = in.nextLong();  
    	long m = in.nextLong();
    	long k = in.nextLong();
 
    	long toAdd;
    	
    	long result = 0;
    	
    	String powStr = "1";
    	long s = (""+m).length();
    	for(int i = 0; i < s; i++) powStr += "0";
		long pow = Long.parseLong(powStr);
    	
    	while(true){
    		
	    	
    		long cost = s*k;
    		
    		toAdd = w / cost;
    		
    		if(toAdd < 1) break;
    		
    		// int pow = (int) Math.pow(10, s);
    		
    		
    		pow = Long.parseLong(powStr);
    		
    		
    		if(m + toAdd >= pow){
    			toAdd = pow - m;
    			m = pow;
    			w -= (toAdd * cost);
    			result += toAdd;
    			s++;
    			powStr += "0";
    		}
    		else{
    			m += toAdd;
    			w -= (toAdd * cost);
    			result += toAdd;
    			break;
    		}
    	}
    	
    	System.out.println(result);
    }
    
    // s(n) is the logarithm in base 10
    
}