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 proconio::input;

fn main() {
    input! {
        q: usize,
        lr: [(i64, i64); q],
    };

    let n = *lr.iter().map(|(_, r)| r).max().unwrap() as usize;

    let mut sieve = vec![true; n + 1];
    sieve[0] = false;
    sieve[1] = false;
    for i in 2.. {
        if i * i > n {
            break;
        }
        if sieve[i] {
            for j in (i + i..=n).step_by(i) {
                sieve[j] = false;
            }
        }
    }

    let mut c = vec![0; n + 2];
    for i in 1..=n {
        c[i] = c[i - 1]
            + if i % 2 == 0 {
                0
            } else {
                if sieve[i] && sieve[(i + 1) / 2] {
                    1
                } else {
                    0
                }
            };
    }

    for &(l_i, r_i) in lr.iter() {
        println!("{}", c[r_i as usize] - c[(l_i - 1) as usize]);
    }
}