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.BigInteger;
import java.util.*;

public class Pr373B {
	public static void main(String[] args) throws IOException {
		new Pr373B().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;

	String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out, true);
		solve();
		out.flush();
	}

	long w;
	long[] st10;
	long k;

	int whichTo(long x, int l, int r) {
		if (r == l + 1) {
			return r;
		}
		int m = (l + r) / 2;
		if (st10[m] > x) {
			return whichTo(x, l, m);
		} else {
			return whichTo(x, m, r);
		}
	}

	boolean can(long st, long m) {
		int l = whichTo(st, 0, 17);
		int r = whichTo(st + m - 1, 0, 17);

		if (l == r) {
			BigInteger i = new BigInteger(m + "");
			i = i.multiply(BigInteger.valueOf(k));
			i = i.multiply(BigInteger.valueOf(l));

			return i.compareTo(BigInteger.valueOf(w)) <= 0;
		}
		BigInteger ans = BigInteger.valueOf(st10[l] - st);
		ans = ans.multiply(BigInteger.valueOf(k));
		ans = ans.multiply(BigInteger.valueOf(l));
		ans = ans.add(BigInteger.valueOf(st + m - st10[r - 1])
				.multiply(BigInteger.valueOf(k))
				.multiply(BigInteger.valueOf(r)));
		for (int i = l + 1; i < r; i++) {
			ans = ans.add(BigInteger.valueOf(st10[i] - st10[i - 1])
					.multiply(BigInteger.valueOf(k))
					.multiply(BigInteger.valueOf(i)));
		}
		return ans.compareTo(BigInteger.valueOf(w)) <= 0;
	}

	long binSearch(long st, long l, long r) {
		if (r == l + 1) {
			return l;
		}
		long m = (l + r) / 2;
		if (can(st, m)) {
			return binSearch(st, m, r);
		} else {
			return binSearch(st, l, m);
		}
	}

	void solve() throws IOException {
		w = nextLong();
		st10 = new long[17];
		st10[0] = 1;
		for (int i = 1; i < 17; i++) {
			st10[i] = st10[i - 1] * 10;
		}
		long m = nextLong();
		k = nextLong();
		out.println(binSearch(m, 0, w + 1));
	}
}