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
s=input()
a=s.split()
count1=0
sum=0
a[0]=int(a[0])
a[1]=int(a[1])
a[2]=int(a[2])
z=str(a[1])
count1=len(z)
print(count1)
z=pow(10,count1)
z=z-a[1]
if z*count1*a[2]>=a[0]:
    print(int(a[0]/(a[2]*count1)))
    exit()
else:
    a[0]=a[0]-z*count1*a[2]
    sum=sum+z
    while 1:
        l=pow(10,count1)
        count1=count1+1
        z=pow(10,count1)
        if a[0]-(z-l)*count1*a[2]>0:
            a[0]=a[0]-(z-l)*count1*a[2]
            sum=sum+z-l
        else:
            sum=sum+a[0]/(a[2]*count1)
            break
sum=int(sum)
print(sum)