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
//373Bmaking sequence is fun
import java.io.*;
import java.util.*;
public class B
{
    public long digit(long x)
{
long e=0;
while(x>0)
{
x=x/10;
e++;}
return e;
}
public static void main(String args[])throws IOException
{
Scanner sc=new Scanner(System.in);
long w=sc.nextLong();
long m=sc.nextLong();
long k=sc.nextLong();
long v=0;
B d=new B();
int cx=0;
long x=d.digit(m);
//System.out.println(w+" "+m+" "+k+" "+x);
while(cx!=1)
{long z=(long)((long)Math.pow(10,x)-m);
//System.out.println(z);
if((w/z)>=(k*x))
{w-=k*x*z;
v=v+z;
m=(long)Math.pow(10,x);
x++;
}
else
{
long b=(long)((long)w/(long)(k*x));
v+=b;
cx=1;
break;
}
}
System.out.println(v);
}
}
