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 log

w, m, k = map(int, input().split())

def calcular_sumatoria(n):
    if n == 0:
        return 0
    suma = 0
    indice = 0
    log_10 = int(log(n, 10))
    if log_10 == 0:
        return n
    for i in range(log_10):
        suma += (i+1) * (10 ** (i+1)) - 10 ** i
    queda = n - 10 ** log_10 + 1
    suma += (log_10 + 1) * queda
    return suma

ope = w/k + calcular_sumatoria(m-1)
# print(ope)
s = m
while ope >= calcular_sumatoria(s):
    s += 1
print(s - m)
  			   	  		 	 		 		  	 	 	 		