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.util.*;
import java.io.*;

public class B373 implements Runnable {
		
	public void solve() throws IOException {
		long money = nextLong();
                long start = nextLong();
                long costPerDigit = nextLong();
                
                long digitsICanAdd = money / costPerDigit;
                
                long digitsBefore = count(start - 1);
                long lo = 0;
                long hi = (long)(1e+17);
                while(hi - lo > 1){
                    long mid = (hi + lo) / 2;
                    long curCountDigits = count(mid);
                    if(curCountDigits - digitsBefore <= digitsICanAdd) lo = mid;
                    else hi = mid;
                }
                
                
                long travelled  = lo - start + 1;
                if(travelled <= 0)
                    out.println(0);
                else
                    out.println(travelled);
	}
        
        
        private long count(long end){
            int len = 1;
            long countDigits = 0;
            long has = 9;
            for(len = 1; len <= 16; len++){
                if(end >= has){
                    countDigits += (len * has);
                    end -= has;
                    has = has * 10;
                }
                else
                    break;
            }
            
            countDigits += end * len;
            return countDigits;
        }

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new B373().run();
	}

        public void debug(Object... arr){
            System.out.println(Arrays.deepToString(arr));
        }

        public void print1Int(int[] a){
                for(int i = 0; i < a.length; i++)
                        System.out.print(a[i] + " ");
                System.out.println();
        }
        
        public void print2Int(int[][] a){
                for(int i = 0; i < a.length; i++){
                        for(int j = 0; j < a[0].length; j++){
                                System.out.print(a[i][j] + " ");
                        }
                        System.out.println();
                }
        }
        
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
                        out = new PrintWriter(System.out);
			tok = null;
			solve();
			in.close();
                        out.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

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

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

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

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
        PrintWriter out;
	BufferedReader in;
	StringTokenizer tok;
}