Recently, the bear started studying data structures and faced the following problem.

You are given a sequence of integers *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*n*</sub> of length *n* and *m* queries, each of them is characterized by two integers *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>. Let's introduce *f*(*p*) to represent the number of such indexes *k*, that *x*<sub class="lower-index">*k*</sub> is divisible by *p*. The answer to the query *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> is the sum: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5e448f6c1a0e40843cecdbbeb95970464626032f.png" style="max-width: 100.0%;max-height: 100.0%;"/>, where *S*(*l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>) is a set of prime numbers from segment [*l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>] (both borders are included in the segment).

Help the bear cope with the problem.
The first line contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">6</sup>). The second line contains *n* integers *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*n*</sub> (2 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">7</sup>). The numbers are not necessarily distinct.

The third line contains integer *m* (1 ≤ *m* ≤ 50000). Each of the following *m* lines contains a pair of space-separated integers, *l*<sub class="lower-index">*i*</sub> and *r*<sub class="lower-index">*i*</sub> (2 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ 2·10<sup class="upper-index">9</sup>) — the numbers that characterize the current query.
'''
    Codeforces 385B - Bear and prime numbers
    29/01/2014
'''
LIM = 10**7

#n = numero massimo fino a cui cerca i primi
#l[i]=True sse i è primo
#pri = lista dei primi fino a n
def eratostene( n ):
    l = [True for x in range(n+1)]
    pri = []
    l[0] = l[1] = False
    i=2
    while i<n+1:
        if l[i]==True:
            pri.append( i )
            j=i*2
            while j<n+1:
                l[j] = False
                j += i
        i+=1
    return (l, pri)
    
_, pri = eratostene( LIM )

n = int(input())
l = [int(x) for x in input().split()]
count = [0 for x in range( LIM+1 )]
for x in l:
    count[x] += 1

f = []
pre = [0 for x in range(LIM+1)]
for x in pri:
    y=x
    s=0
    while y<=LIM:
        s+=count[y]
        y+=x
    f.append(s)
    pre[x] = s

for i in range(1,LIM+1):
    pre[i] += pre[i-1]    
        
m = int(input())
for x in range(m):  
    l, r = [int(x) for x in input().split()]
    if l>LIM:
        l=LIM
    if r>LIM:
        r=LIM
    
    print(pre[r]-pre[l-1])
    
        
        
    
    
    
    
    
    
    
    
            
        
        
        
        
        
     
