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 solve(w, m, k):
    # recherche du nombre de chiffres
    ndigits = len(str(m))
    nextm = 10**ndigits
    length = 0
    while True:
        cost = k*ndigits*(nextm-m)
        if cost>w: break
        w -= cost
        length += nextm-m
        m = nextm
        nextm *= 10
        ndigits += 1
    # recherche dichotomique dans [0,nmax[ (nombre de termes à ajouter)
    nmax = nextm+1-m
    nmin = 0
    while nmax-nmin>1:
        cur = (nmin+nmax)//2
        if k*ndigits*cur<=w:
            nmin = cur
        else:
            nmax = cur
    return length+nmin

w, m, k = [int(_) for _ in input().split()]
print(solve(w, m, k))
    
