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.
def clc(): 
    n = int(input()) 
    arr = list(map(int,input().split()))
    prefix = [0]*(pow(10,6+1))
    import math
    def preprocess(arr):

        def findfactor(x): 
            vis = set()
            while x%2 == 0:
                vis.add(2)
                x= x//2 
            for i in range(3,math.ceil(math.sqrt(x))+1,2):
                while x%i == 0:
                    vis.add(i)
                    x= x//i
                
            if x>2:
                vis.add(x)
            return vis

        for ele in arr:
            calc = findfactor(ele)
            for x in calc:
                prefix[x]+=1 
        

        
    preprocess(arr)
    #print(prefix)
    for i in range(1,len(prefix)):
        prefix[i]+=prefix[i-1]
    q = int(input())
    for _ in range(q):
        l,r = map(int,input().split())
    
        ans = prefix[r]-prefix[l-1]
        print(ans)
        

    return True
cc = clc()

if not cc :
    print(0)
   