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.
def log10_(a):
    ans = 0
    while a//10:
        ans += 1
        a = a//10
    return ans

def pow10(a):
    ans = 1
    for i in range(a): ans *= 10
    return ans

w, m, k = [int(x) for x in input().split()]
ans = 0
costo = k*(log10_(m)+1)
for i in range(log10_(m)+1, 18):
    if (pow10(i)-m)*costo > w:
        ans += w//costo
        break
    w -= (pow10(i)-m)*costo
    ans += pow10(i)-m
    costo += k
    m = pow10(i)
print (ans)
	 			 				 	 			 		 	 	 	 				