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.
The first line contains three integers *w* (1 ≤ *w* ≤ 10<sup class="upper-index">16</sup>), *m* (1 ≤ *m* ≤ 10<sup class="upper-index">16</sup>), *k* (1 ≤ *k* ≤ 10<sup class="upper-index">9</sup>).

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.
from math import *
def S(n):
	return floor(log10(n)) + 1

def sumTo(x,m):
	leng = S(m)
	suma = 0
	done = False
	while (not done):
		q = 0
		if(leng == S(x)):
			q = x - 10**(leng-1)
			done = True
		elif(leng == S(m)):
			q = 10**(leng) - m
		else:
			q = 10**(leng) - 10**(leng-1)
		suma+=q*k*leng
		leng+=1
	return suma

inp = input()
w, m, k = [int(x) for x in inp.split(" ")]

L = m
R = int(5e17)
mid = 0
while(L < R):
	mid = int((L+R)/2)
	if(sumTo(mid,m) > w):
		R = mid
	else:
		L = mid+1
print(L-m-1)

 	  			 	 			 			 	     		 	 		