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.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;


public class sequence {
	private static BigInteger maxNum(int len)
	{
		String val = "";
		for(int i=0;i<len;++i)
		{
			val+="9";
		}
		return new BigInteger(val);
	}
public static void main(String[] args) throws Exception {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));

	StringTokenizer st = new StringTokenizer(br.readLine());
	BigInteger NINE = new BigInteger("9");
	BigInteger w = new BigInteger(st.nextToken()),m = new BigInteger(st.nextToken()) ;
	long k = Long.parseLong(st.nextToken());
	BigInteger count = BigInteger.ZERO;
	int mLen = String.valueOf(m).length();
	BigInteger val = maxNum(mLen);
	while(w.compareTo(BigInteger.ZERO)>0)
	{
	BigInteger remaining = val.subtract(m).add(BigInteger.ONE);
	BigInteger costOfOne = new BigInteger(mLen*(long)k+"");
	BigInteger totalCost = remaining.multiply(costOfOne);
	BigInteger toSpend = w.subtract(totalCost);
	if(toSpend.compareTo(BigInteger.ZERO)<0)
	{
		remaining = w.divide(costOfOne);
		count=count.add(remaining);
		break;
	}
	else
	{
		count=count.add(remaining);
		w=w.subtract(totalCost);
	}
	mLen++;
	m = val.add(BigInteger.ONE);
	val=(val.multiply(BigInteger.TEN)).add(NINE);
	}
	pw.println(count);
	pw.flush();
	pw.close();
}
}
