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 std::io::*;
use std::str::FromStr;

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let input_max = 100_000;
    let mut is_primes = [true; 100_000 + 10];
    let mut like2017 = [false; 100_000 + 10];
    let mut accum_sum = [0; 100_000 + 10];
    let mut primes = Vec::new();

    is_primes[0] = false;
    is_primes[1] = false;

    //前処理--------------------------
    //エラトステネスの篩で素数を列挙
    for i in 2..=input_max {
        if is_primes[i] {
            primes.push(i);
            for j in (2 * i..input_max + 1).step_by(i) {
                is_primes[j] = false;
            }
        }
    }

    // 2017に似た数をチェックし累積和にする
    for i in 2..=input_max {
        if is_primes[i] && is_primes[(i + 1) / 2] {
            like2017[i] = true;
            accum_sum[i] = accum_sum[i - 1] + 1;
        } else {
            accum_sum[i] = accum_sum[i - 1];
        }
    }

    //クエリ処理（入力を受ける）--------
    let q: usize = read();
    for _ in 0..q {
        let l: usize = read();
        let r: usize = read();

        println!("{}", accum_sum[r] - accum_sum[l - 1]);
    }

    /*
    for i in 0..17{
        println!("like2017:{}",like2017[i]);
        println!("accum_sum:{}",accum_sum[i]);
    }
    */
}