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.
from sys import *
from bisect import *
from collections import *

INF = 1e9
Input = []
MOD = 1000000007

stdin = open('in', 'r')
#stdout = open('out', 'w')

def Out(x):
    stdout.write(str(x) + '\n')

def In():
    return stdin.readline().strip()

def inputGrab():
    for line in stdin:
        Input.extend(map(str, line.strip().split()))

############################################################

isPrime = [True for i in range(7100)]
primes = []
divCount = [0 for i in range(10000100)]

def sieve(lim = 7000):
    global isPrime
    global primes
    
    isPrime[0] = isPrime[1] = 0
    for i in range(2, lim):
        if isPrime[i]:
            primes.append(i)
            for j in range(i*i, lim, i):
                isPrime[j] = 0


def factorize(val, Cnt):
    global primes
    global divCount
    
    for prime in primes:
        if val < prime:
            break
        if val % prime == 0:
            divCount[prime] += Cnt
        while val >= prime and val % prime == 0:
            val /= prime
    
    if val != 1:
        divCount[val] += Cnt



def main():
    n = int(In())
    #v = map(int, In().split())
    global divCount
    global primes
    
    sieve()
        
    cnt = {}
    for elem in map(int, In().split()):
        if elem in cnt:
            cnt[elem] += 1
        else:
            cnt[elem] = 1
    
    #print("*")
    for elem, Cnt in cnt.items():
        factorize(elem, Cnt)
        
    #print("done")
    # Cumulative frequency
    for i in range(1, 10000001):
        divCount[i] += divCount[i-1]
        
    q = int(In())
    
    for i in range(q):
        l, r = map(int, In().split())
        r = min(10000000, r)
        print(divCount[r]-divCount[l-1])
        
        
    
if __name__ == '__main__':
    main()
