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 *a*<sub class="lower-index">*i*</sub> 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 *c*<sub class="lower-index">*j*</sub> 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 10<sup class="upper-index">9</sup> + 7.
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 *a*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 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 *c*<sub class="lower-index">*j*</sub> (1 ≤ *c*<sub class="lower-index">*j*</sub> ≤ 1 000 000), the number of flowers of a new species Farmer John receives on day *j*.
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.
3 3 2
4
6
9
8
6
5
16
4 1 2
6
5
4
3
2
1
20
21
3 3 3
6
8
10
12
14
16
8
20
42
1 1 1
1
1
2
10 10 10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
110
660
2860
10010
30030
80080
194480
437580
923780
1847560
7 1 9
3
6
7
2
10
8
1
3
9
2
9
10
2
4
5
2
40
49
51
60
70
72
76
81
83
5 1 7
10
8
4
5
8
3
9
3
9
3
10
1
38
47
50
59
62
72
73
7 2 8
2
5
7
9
6
2
5
2
8
7
10
8
8
5
7
40
52
67
90
113
144
169
195
9 8 10
10
9
6
8
8
8
8
6
2
7
1
9
5
10
8
1
1
6
10
46
166
496
1288
3012
6480
12915
24355
43923
76077
from sys import stdin #Bibliotecas para optimizar la entrada/salida
from sys import stdout

MOD = 1000000007 #Modulo
MAXA = 1000007 #Maximo valor posible en A
MAXN = 200007  #Suma de maximos valores posibles de n y q

p = [0 for i in range(MAXA)]      #Para cada i me guarda 0 si i es primo o 1 en otro caso
phi = [i for i in range(MAXA)]   #Para cada i conserva phi(i)
cant = [0 for _ in range(MAXA)]  #Guarda por cada indice j la cantidad de valores en A_i que son multiplos de j
combin = [0 for _ in range(MAXA)] #Guarda por cada indice j combinatoria de cant[j] en k

#Computamos la funcion phi para cada numero
def ComputePhi():
    p[1] = p[0] = 1
    for i in range(1,MAXA): #Pasamos por cada numero
        for j in range(i,MAXA,i): #y por cada uno de sus multiplos
            if p[i] == 0:   #Si el numero es primo damos un paso en la computacion de phi de su multiplo
                if i != j: 
                    p[j] = 1    #Su multiplo no sera primo
                phi[j] = (phi[j]//i)*(i-1)
            if dt.get(j) != None:   #Si este multiplo esta en A_q
                div[dt[j]].append(i)  #Guardo i como uno de sus divisores (al final va a tener todos sus divisores)

def qp(a,p): #Exponenciacion binaria O(log p)
    a = int(a)
    p = int(p)
    if p == 0:
        return 1
    x = qp(a,p//2)
    x = (x*x)%MOD
    if p%2 == 1:
        x = (x*a)%MOD
    return x

def inv(a): #inverso multiplicativo, ya que el modulo es primo usamos el pequeño teorema de fermat para hallarlo
    return qp(a,MOD-2)

inputs = list(map(int,stdin.readline().split())) #lee la entrada
n,k,q= inputs[0],inputs[1],inputs[2]  

fact = [0 for i in range(MAXN)] #Lista que va a mantener para cada valor entre 0 y MAXN el factorial de ese numero (modulado)
ifact = [0 for i in range(MAXN)] #Lista que va a mantener para cada valor entre 0 y MAXN el inverso multiplicativo del factorial de ese numero.

#Coeficientes binomiales:
def comb(a,b): 
    if a-b < 0:
        return 0
    return ((fact[a]*ifact[b])%MOD*ifact[a-b])%MOD

#Computamos el factorial recursivamente:
fact[0] = 1
for i in range(1,MAXN):
    fact[i] = (fact[i-1]*i)%MOD

#Computamos el inverso multiplicativo del factorial recursivamente teniendo en cuenta que (k!)^(-1) = ((k+1)!^(-1))*k+1
ifact[MAXN-1] = inv(fact[MAXN-1])
for i in range(MAXN-2,-1,-1):
    ifact[i] = (ifact[i+1]*(i+1))%MOD

dt = {} #Diccionario que mapea cada numero en A_q diferente a una posicion diferente en el array div
ln = [] #Lista que guarda $A$
c = 0
for j in range(n):
    a = int(stdin.readline())
    if dt.get(a) == None:
        c+=1
        dt[a] = c
    ln.append(a)
lq = [] #Lista que guarda todos los w_i
for _ in range(q):
    a = int(stdin.readline())
    if dt.get(a) == None:
        c+=1
        dt[a] = c
    lq.append(a)

#El mismo guardara los divisores de cada numero en A_q
div = [[] for _ in range(c+1)]  #Crear el array div con tantas posiciones como numeros diferentes en A_q
ComputePhi()

for a in ln:
    for i in div[dt[a]]:
        cant[i]+=1  #por cada divisor x de a suma 1 en la posicion x
ans = 0
for i in range(1,MAXA): #Calcula la respuesta para A (ver informe)
    combin[i] = comb(cant[i],k)
    ans += (phi[i]*combin[i])%MOD
    ans%=MOD

for a in lq: #por cada w_i calcula la respuesta para A_i...
    for i in div[dt[a]]:   #Pasando por cada divisor de w_i
        ans -= (combin[i]*phi[i])%MOD
        ans%=MOD
        if ans < 0:
            ans+=MOD
        cant[i]+=1  #Mantenemos por limpieza actualizado los valores de cant
        combin[i] = comb(cant[i],k) #y de combin
        ans += (combin[i]*phi[i])%MOD
        ans%=MOD
    print(ans)