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.math.*;

public class Cfthree {
	public static long w , m , k , num;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while ( sc.hasNextLong() ) {
			w = sc.nextLong();
			m = sc.nextLong();
			k = sc.nextLong();
			num = calDig(m);
			long L = 0;
			long R = w / k + 10;
			long res = 0;
			while ( L<=R ) { //目的是得到 尽量多的个数
				long M = ( L + R ) >> 1;
				//System.out.println("zzzz "+L+" "+" "+R+" "+M);
				if ( bSearch(M) ) {
					res = M;
					L = M + 1;
				} else {
					R = M - 1;
				}
			}
			System.out.println(res);
		}
	}
	public static boolean bSearch( long M ) { // M是表示多少个数
		long var = num;
		long now = m;
		long sum = 0;
		while ( M>0 ) {
			long cnt = calPow(10,var) - now;
			long temp = Math.min(cnt, M);
			M -= temp;
			//System.out.println("++++++ "+cnt+" "+temp+" "+M);
			sum += var * temp * k;
			if ( sum>w ) {
				//System.out.println("\\\\\\\\"+" "+sum+" "+w);
				return false;
			}
			now = calPow(10,var);
			++ var;
		}
		//System.out.println("-------"+" "+sum+" "+w);
		return sum <= w;
	}
	public static long calPow( long x , long y ) {
		long res = 1;
 		while ( y>=1 ) {
 			res *= x;
 			-- y;
 		}
 		return res;
	}
	public static long calDig( long x ) {
		long num = 0;
		while ( x>0 ) {
			x /= 10;
			++ num;
		}
		return num;
	}
}
/*
462 183 8
*/