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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class B {

	public static void main(String[] args) throws IOException {
		FastScanner scan = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		long w = scan.nextLong(), m = scan.nextLong(), k = scan.nextLong();
		BigInteger lo = BigInteger.ZERO, hi = new BigInteger("1000000000000000000"), max = BigInteger.ZERO;
		while(lo.compareTo(hi) == -1){
			BigInteger mid = lo.add(hi).divide(new BigInteger("2"));
			if(possible(mid, w, m, k)){
				max = mid;
				lo = mid.add(BigInteger.ONE);
			} else hi = mid;
		}
		out.println(max.subtract(BigInteger.valueOf(m)).add(BigInteger.ONE));
		out.close();
	}
	
	static boolean possible(BigInteger mid, long w, long m, long k){
		return cost(mid, k).subtract(cost(BigInteger.valueOf(m-1), k)).compareTo(BigInteger.valueOf(w)) != 1;
	}
	
	static BigInteger cost(BigInteger n, long k){
		if(n.compareTo(BigInteger.ZERO) == 0) return BigInteger.ZERO;
		String a = n.toString();
		String b = "1";
		for(int i = 1; i < a.length(); i++) b += "0";
		return ((n.subtract(new BigInteger(b)).add(BigInteger.ONE))).multiply(BigInteger.valueOf(a.length())).multiply(BigInteger.valueOf(k)).add(cost(new BigInteger(b).subtract(BigInteger.ONE), k));
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

		public int nextInt() {return Integer.parseInt(next());}

		public long nextLong() {return Long.parseLong(next());}

		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			try {line = br.readLine();} 
			catch (Exception e) {e.printStackTrace();}
			return line;
		}

		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		
		public long[] nextLongArray(int n){
			long[] a = new long[n];
			for(int i = 0; i < n; i++) a[i] = nextLong();
			return a;
		}
		
		public double[] nextDoubleArray(int n){
			double[] a = new double[n];
			for(int i = 0; i < n; i++) a[i] = nextDouble();
			return a;
		}
		
		public char[][] nextGrid(int n, int m){
			char[][] grid = new char[n][m];
			for(int i = 0; i < n; i++) grid[i] = next().toCharArray();
			return grid;
		}
	}

}
