Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

C. Bear and Prime Numbers

time limit per test

2 secondsmemory limit per test

512 megabytesinput

standard inputoutput

standard outputRecently, the bear started studying data structures and faced the following problem.

You are given a sequence of integers *x*_{1}, *x*_{2}, ..., *x*_{n} of length *n* and *m* queries, each of them is characterized by two integers *l*_{i}, *r*_{i}. Let's introduce *f*(*p*) to represent the number of such indexes *k*, that *x*_{k} is divisible by *p*. The answer to the query *l*_{i}, *r*_{i} is the sum: , where *S*(*l*_{i}, *r*_{i}) is a set of prime numbers from segment [*l*_{i}, *r*_{i}] (both borders are included in the segment).

Help the bear cope with the problem.

Input

The first line contains integer *n* (1 ≤ *n* ≤ 10^{6}). The second line contains *n* integers *x*_{1}, *x*_{2}, ..., *x*_{n} (2 ≤ *x*_{i} ≤ 10^{7}). 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*_{i} and *r*_{i} (2 ≤ *l*_{i} ≤ *r*_{i} ≤ 2·10^{9}) — the numbers that characterize the current query.

Output

Print *m* integers — the answers to the queries on the order the queries appear in the input.

Examples

Input

6

5 5 7 10 14 15

3

2 11

3 12

4 4

Output

9

7

0

Input

7

2 3 5 7 11 4 8

2

8 10

2 123

Output

0

7

Note

Consider the first sample. Overall, the first sample has 3 queries.

- The first query
*l*= 2,*r*= 11 comes. You need to count*f*(2) +*f*(3) +*f*(5) +*f*(7) +*f*(11) = 2 + 1 + 4 + 2 + 0 = 9. - The second query comes
*l*= 3,*r*= 12. You need to count*f*(3) +*f*(5) +*f*(7) +*f*(11) = 1 + 4 + 2 + 0 = 7. - The third query comes
*l*= 4,*r*= 4. As this interval has no prime numbers, then the sum equals 0.

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Apr/23/2018 03:14:58 (d2).

Desktop version, switch to mobile version.

User lists

Name |
---|