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: Wrong Answer)

use std::io::Read;

pub fn main() {
    const MAX: usize = 100001;
    let mut is_prime = [true;MAX];
    is_prime[0] = false;
    is_prime[1] = false;
    for i in 2..MAX {
        if !is_prime[i] {continue;}
        let mut m = 2;
        while i*m < MAX {
            is_prime[i*m] = false;
            m += 1;
        }
    }

    let mut is_likenum = [false;MAX];
    for i in 0..MAX {
        if i % 2 == 1 && is_prime[i] && is_prime[(i+1)/2] {
            is_likenum[i] = true;
        }
    }
    
    let mut csum = [0;MAX + 1];
    for i in 0..MAX {
        csum[i+1] = csum[i] + (if is_likenum[i] {1} else {0})
    }

    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let q: u32 = iter.next().unwrap().parse().unwrap();
    for _ in 0..q {
        let left: usize = iter.next().unwrap().parse().unwrap();
        let right: usize = iter.next().unwrap().parse().unwrap();
        println!("{}", csum[right] - csum[left - 1]);
    }
}