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 (1.15.1) (judge verdict: Time Limit Exceeded)

#[allow(unused_macros)]
macro_rules! invec {
    ( $t:ty ) => {{
        let mut s = String::new();
        match std::io::stdin().read_line(&mut s) {
            Ok(0) => Vec::<$t>::new(),
            Ok(n) => s.trim().split_whitespace().map(|s| s.parse::<$t>().unwrap()).collect::<Vec<$t>>(),
            Err(_) => Vec::<$t>::new(),
        }
    }}
}

#[allow(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($(
            {
                 splits.next().unwrap().parse::<$t>().unwrap()
            },
        )*)
    }}
}

struct Primes {
    prime_list: Vec<bool>
}

impl Primes {
    fn new(n: usize) -> Self {
        let mut lis = vec![false; n+1];
        lis[2] = true;

        for i in (3..n+1).filter(|a| a % 2 == 1) {
            let i_sqrt = (i as f64).sqrt().ceil() as usize;
            lis[i] = !(3..i_sqrt+1).fold(false, |flag, now| flag || i % now == 0);
        }

        Primes { prime_list: lis }

    }

    fn is_prime(&self, n: usize) -> bool {
        if (n as usize) < self.prime_list.len() {
            self.prime_list[n as usize]
        } else {
            panic!(n.to_string() + " hasn't been calculated yet.");
        }
    }
}


#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (q,) = input!(usize);
    let p = Primes::new(100000);

    for _ in 0..q {
        let (l, r) = input!(usize, usize);
        println!("{}", (l..r+1).filter(|n| *n % 2 == 1 && p.is_prime(*n) && p.is_prime((*n+1) / 2)).count());
    }


}

fn main() {
    solve();
}