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::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let q: usize = input();
    let lr: Vec<(usize, usize)> = (0..q)
        .map(|_| {
            let l = input();
            let r = input();
            (l, r)
        })
        .collect();

    let mut v = vec![];
    for i in 2..100001 {
        if i % 2 == 0 {
            continue;
        }
        if is_prime(i) && is_prime((i + 1) / 2) {
            v.push(i);
        }
    }

    let mut sum = vec![0i64; 100001];
    for i in 1..100001 {
        sum[i] = sum[i - 1];
        if v.contains(&i) {
            sum[i] += 1;
        }
    }

    for i in 0..q {
        println!("{}", sum[lr[i].1] - sum[lr[i].0 - 1]);
    }
}

fn is_prime(x: usize) -> bool {
    let mut ok = true;
    for i in 2..x {
        if i * i > x {
            break;
        }
        if x % i == 0 {
            ok = false;
            break;
        }
    }
    ok
}