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 SieveOfEratosthenes(limit):
    """Returns all primes not greater than limit."""
    isPrime = [True]*(limit+1)
    isPrime[0] = isPrime[1] = False
    primes = []
    for i in range(2, limit+1):
        if not isPrime[i]:continue
        primes += [i]
        for j in range(i*i, limit+1, i):
            isPrime[j] = False
    return primes
from collections import Counter


def gcd(x, y):
    """greatest common divisor of x and y"""
    while y:
        x, y = y, x % y
    return x


def memodict(f):
    """memoization decorator for a function taking a single argument"""
    class memodict(dict):
        def __missing__(self, key):
            ret = self[key] = f(key)
            return ret

    return memodict().__getitem__


def pollard_rho(n):
    """returns a random factor of n"""
    if n & 1 == 0:
        return 2
    if n % 3 == 0:
        return 3

    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            prev = p
            p = (p * p) % n
            if p == 1:
                return gcd(prev - 1, n)
            if p == n - 1:
                break
        else:
            for i in range(2, n):
                x, y = i, (i * i + 1) % n
                f = gcd(abs(x - y), n)
                while f == 1:
                    x, y = (x * x + 1) % n, (y * y + 1) % n
                    y = (y * y + 1) % n
                    f = gcd(abs(x - y), n)
                if f != n:
                    return f
    return n


@memodict
def prime_factors(n):
    """returns a Counter of the prime factorization of n"""
    if n <= 1:
        return Counter()
    f = pollard_rho(n)
    return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)

primes=SieveOfEratosthenes(10000000)
score=[0]*(10000001)
n=int(input())
a=[int(x) for x in input().split()]
for i in a:
    for j in prime_factors(i):
        score[j]+=1
for i in range(1,len(score)):
    score[i]+=score[i-1]
m=int(input())
for i in range(m):
    l,r=[int(x) for x in input().split()]
    r=min(10000000,r)
    if l<=r:
        print(score[r]-score[l-1])
    else:
        print(0)
    
 