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.
import sys
import math as mt
input=sys.stdin.buffer.readline  
MAXN = 10000001
  
# stores smallest prime factor for 
# every number 
spf = [0 for i in range(MAXN)] 
  
# Calculating SPF (Smallest Prime Factor)  
# for every number till MAXN. 
# Time Complexity : O(nloglogn) 
def sieve(): 
    spf[1] = 1
    for i in range(2, MAXN): 
          
        # marking smallest prime factor  
        # for every number to be itself. 
        spf[i] = i 
  
    # separately marking spf for  
    # every even number as 2 
    for i in range(4, MAXN, 2): 
        spf[i] = 2
  
    for i in range(3, mt.ceil(mt.sqrt(MAXN))): 
          
        # checking if i is prime 
        if (spf[i] == i): 
              
            # marking SPF for all numbers 
            # divisible by i 
            for j in range(i * i, MAXN, i):  
                  
                # marking spf[j] if it is  
                # not previously marked 
                if (spf[j] == j): 
                    spf[j] = i 

t=1
sieve()
#print(len(spf))
pref=[0]*(MAXN)
#print(len(pref))
for __ in range(t):
    n=int(input())
    x=list(map(int,input().split()))
    for i in range(n):
        x1=x[i]
        while x1!=1:
            pref[spf[x1]]+=1
            x1=x1//spf[x1]
    m=int(input())
    for i in range(1,MAXN):
        pref[i]+=pref[i-1]
    for ____ in range(m):
        xi,yi=map(int,input().split())
        print(pref[yi]-pref[xi-1])
     
	