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::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    const MAX: usize = 100_000;
    let primes = enumerate_primes(MAX);
    let mut s: Vec<usize> = vec![0; MAX + 1];
    let mut i = 3;
    while i <= MAX {
        if primes.contains(&i) && primes.contains(&((i + 1) / 2)) {
            s[i] = s[i - 2] + 1;
        } else {
            s[i] = s[i - 2];
        }
        i += 2;
    }

    let q: usize = iterator.next().unwrap().parse().unwrap();
    let mut result = String::new();
    for _ in 0..q {
        let l: usize = iterator.next().unwrap().parse().unwrap();
        let r: usize = iterator.next().unwrap().parse().unwrap();

        let sl = if l == 1 {
            0
        } else {
            s[l - 2]
        };
        let sr = s[r];
        result.push_str(&format!("{}\n", sr - sl));
    }
    return result.trim().to_string();
}

fn enumerate_primes(n: usize) -> std::collections::HashSet<usize> {
    let mut result = std::collections::HashSet::new();
    for i in 2..(n + 1) {
        let mut is_prime = true;
        let mut x = 2;
        while x * x <= i {
            if i % x == 0 {
                is_prime = false;
                break;
            }
            x += 1;
        }
        if is_prime {
            result.insert(i);
        }
    }
    return result;
}