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 stdin
put = lambda : list(map(int, stdin.readline().split()))


def sieve(N):
    isPrime = [True]*N
    isPrime[0] = isPrime[1] = False
    leastPrime = list(range(N))
    for i in range(2, N):
        if isPrime[i]:
            for j in range(i*i, N, i):
                if isPrime[j]:
                    isPrime[j] = False
                    leastPrime[j] = i
    return leastPrime


n, = put()
l = put()
m, = put()
N = int(1e7+1)
count = [0]*N
leastPrime = sieve(N)

for i in l:
    while i!=1:
        lp = leastPrime[i]
        count[lp]+=1
        while leastPrime[i]==lp:
            i//=lp

for i in range(2, N):
    count[i]+=count[i-1]
        
ans = []
for _ in range(m):
    l,r = put()
    l = min(N-1, l-1)
    r = min(N-1, r)
    ans.append(count[r]-count[l])
   
print(*ans)
#print(*count[:20])