Cowslip Collections
Contest: CROC 2016 - Elimination Round
Difficulty rating: 2500
Tags: combinatorics, math, number theory

In an attempt to make peace with the Mischievious Mess Makers, Bessie and Farmer John are planning to plant some flower gardens to complement the lush, grassy fields of Bovinia. As any good horticulturist knows, each garden they plant must have the exact same arrangement of flowers. Initially, Farmer John has n different species of flowers he can plant, with ai flowers of the i-th species.

On each of the next q days, Farmer John will receive a batch of flowers of a new species. On day j, he will receive cj flowers of the same species, but of a different species from those Farmer John already has.

Farmer John, knowing the right balance between extravagance and minimalism, wants exactly k species of flowers to be used. Furthermore, to reduce waste, each flower of the k species Farmer John chooses must be planted in some garden. And each of the gardens must be identical; that is to say that each of the k chosen species should have an equal number of flowers in each garden. As Farmer John is a proponent of national equality, he would like to create the greatest number of gardens possible.

After receiving flowers on each of these q days, Farmer John would like to know the sum, over all possible choices of k species, of the maximum number of gardens he could create. Since this could be a large number, you should output your result modulo 109 + 7.

Input format:
The first line of the input contains three integers n, k and q (1 ≤ k ≤ n ≤ 100 000, 1 ≤ q ≤ 100 000).

The i-th (1 ≤ i ≤ n) of the next n lines of the input contains an integer ai (1 ≤ ai ≤ 1 000 000), the number of flowers of species i Farmer John has initially.

The j-th (1 ≤ j ≤ q) of the next q lines of the input contains an integer cj (1 ≤ cj ≤ 1 000 000), the number of flowers of a new species Farmer John receives on day j.

Output format:
After each of the q days, output the sum of the maximum possible number of gardens, where the sum is taken over all possible choices of k species, modulo 109 + 7.

Constraints:
Time limit: 8.0 s
Memory limit: 512.0 MB

Note:
In the first sample case, after the first day Farmer John has (4, 6, 9, 8) of each type of flower, and k = 3.

Choosing (4, 6, 8) lets him make 2 gardens, each with (2, 3, 4) of each flower, respectively. Choosing (4, 6, 9), (4, 9, 8) and (6, 9, 8) each only let him make one garden, since there is no number of gardens that each species can be evenly split into. So the sum over all choices of k = 3 flowers is 2 + 1 + 1 + 1 = 5.

After the second day, Farmer John has (4, 6, 9, 8, 6) of each flower. The sum over all choices is 1 + 2 + 2 + 1 + 1 + 2 + 2 + 3 + 1 + 1 = 16.

In the second sample case, k = 1. With x flowers Farmer John can make x gardens. So the answers to the queries are 6 + 5 + 4 + 3 + 2 = 20 and 6 + 5 + 4 + 3 + 2 + 1 = 21.

Example input:
3 3 2
4
6
9
8
6

Example output:
5
16

Example input:
4 1 2
6
5
4
3
2
1

Example output:
20
21

Editorial:
Idea: [user:desert97]
Translating the problem, it is asking for $$\sum_{s} \gcd(S)$$ where $$\gcd(S)$$ denotes the $${ \mathrm { g c d } }$$ of all elements of S, and the sum runs over all subsets of size k or {a1, a2, ..., an}. Let d[i] be the number of indices j such that i|aj.
Let M = 106. I claim that the previous sum is $$\sum_{i=1}^{M} \binom{d[i]}{k} \cdot \phi(i)$$
Proof: Note that $${ \binom { d [ i ] } { k } }$$ counts the number of subsets S of size k such that $$i \mid \gcd(S).$$ Now note that $$\sum_{i \mid n} \phi(i) = n$$, which is a well-known fact. Therefore, if we instead add φ(i) for each subset where $$i \mid \gcd(S)$$, we get i overall for sets where $$i = \gcd(S)$$, as we wanted. Now just implement the above formula to finish.
This can be done in time $$O(M \log M + \sum \tau(a_i))$$, since we have to update all divisors of ai for all i. $$\tau(a_i) \leq 240$$ always, so this is fast.
In a more general sense, we can compute $$\sum_{S} f(\gcd(S))$$, for any function $$f : \mathbb{Z}_{>0} \to \mathbb{Z}_+$$ Then to compute the answer we would just sum $$\sum_{i=1}^{M} \binom{d[i]}{k} \cdot g(i)$$, where g = μ * f, where μ denotes the Möbius function, and  *  denotes Dirichlet convolution. In this problem f(n) = n, and φ = μ * n.