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.
import math

def digitos(x):
    contador=0
    while x>0:
        x//=10
        contador+=1
    return contador

def Secuencia(input):
    secuencia=0
    s = [int(x) for x in input.split(" ")]
    w=s[0]
    m=s[1]
    k=s[2]
    contadorBase = 10 ** digitos(m)
    want= -1

    while want != w:
        want=w
        sobrantes = contadorBase - m
        costo=digitos(m)*k
        if(w-sobrantes*costo >=0):
            w-=sobrantes*costo
            secuencia += sobrantes
            m+=sobrantes
            contadorBase*=10
        

    while w>=0:
        costo= digitos(m)* k
        puedoComprar = w//costo
        w-=puedoComprar*costo
        secuencia+= puedoComprar
        m+=puedoComprar

        if (w-digitos(m)*k<0):
            break
        #w=w-digitos(m)*k
        #secuencia+=1;
        #m=m+1

    return secuencia

def main():


    print(digitos(9))
    inp = input()
    res = Secuencia(inp)
    print(res)


if __name__ == '__main__':
	main()
 	  	 		 		 	 	 										 	 		