Problem:
Score : 400 points

Problem Statement

We say that a odd number N is similar to 2017 when both N and (N+1)/2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Constraints

1≤Q≤10^5

1≤l_i≤r_i≤10^5

l_i and r_i are odd.

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2

Solution in Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
	input! {
		q: usize,
		queries: [(usize, usize); q],
	}

	let mut counts = vec![0; 100001];
	let mut prime_set = vec![false; 100001];
	prime_set[2] = true;
	let mut primes: Vec<usize> = vec![2];
	for i in (3..100000).step_by(2) {
		if primes.iter().all(|&q| i % q != 0 ) {
			primes.push(i);
			prime_set[i] = true;
			if prime_set[(i+1)/2] {
				counts[i] = 1;
			}
		}
	}

	for i in 0..100000 {
		counts[i+1] += counts[i];
	}

	for (l, r) in queries {
		println!("{}", counts[r] - counts[l-1]);
	}
}