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


def size(a):
    return len(str(a))

def how_many(m, n):
    len1 = size(m)
    len2 = size(n)
    if len1 == len2:
        return (n - m + 1) * len1
    else:
        cnt = 0
        for len in range(len1, len2 + 1):
            if len == len1:
                cnt += (pow(10, len1) - m) * len
            elif len == len2:
                cnt += (n - pow(10, len2 - 1)) * len
            else:
                cnt += (pow(10, len) - pow(10, len - 1)) * len
        return cnt + len2

def check(n):
    return how_many(m, n) * k <= w


l = m
r = 10 ** 18
while (r - l > 1):
    n = l + (r - l) // 2
    if check(n):
        l = n
    else:
        r = n
if l != m:
    print(l - m + 1)
else:
    print(int(check(m)))