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.math.*;
import java.util.*;

public class Main {
	public static class Segment
	{
		long l,r;
		Segment(long l, long r)
		{
			this.l=l;
			this.r=r;
		}
		Segment intersect(Segment o)
		{
			return new Segment(Math.max(l,o.l), Math.min(r,o.r));
		}
		boolean isEmpty()
		{
			return l>r;
		}
		long length()
		{
			return isEmpty()?0:r-l+1;
		}
	}
	public static void main(String args[])
	{
		Scanner in=new Scanner(System.in);
		PrintWriter out=new PrintWriter(System.out);
		
		long w,m,k;
		w=in.nextLong();
		m=in.nextLong();
		k=in.nextLong();
		
		long lo=m-1;
		long hi=w+m;
		while(lo<hi)
		{
			long mid=(lo+hi+1)/2;
			long p1=1;
			long p2=9;
			long cost=0;
			
			Segment cur=new Segment(m,mid);
			for(int i=1;i<=17;i++)
			{
				cost+=i*(cur.intersect(new Segment(p1,p2)).length());
				p1*=10;
				p2*=10; p2+=9;
			}
			if(cost<=w/k)
				lo=mid;
			else
				hi=mid-1;
		}
		out.println(lo-m+1);
		out.flush();
	}
}
