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: Accepted)

use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let q: usize = itr.next().unwrap().parse().unwrap();
    let mut out = Vec::new();

    let mut is_prime: Vec<bool> = vec![true; 100010];
    let mut cumsum: Vec<u64> = vec![0; 100010];
    is_prime[0] = false;
    is_prime[1] = false;
    for i in 2..100005 {
        if is_prime[i] {
            let mut j = i * 2;
            while j < 100005 {
                is_prime[j] = false;
                j += i;
            }
        }
    }
    for i in 1..100005 {
        if is_prime[i] && is_prime[(i + 1) >> 1] {
            cumsum[i] += 1;
        }
    }
    for i in 0..100005 {
        cumsum[i + 1] += cumsum[i];
    }

    for _ in 0..q {
        let l: usize = itr.next().unwrap().parse().unwrap();
        let r: usize = itr.next().unwrap().parse().unwrap();
        writeln!(out, "{}", cumsum[r] - cumsum[l - 1]).unwrap();
    }
    stdout().write_all(&out).unwrap();
}