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)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    dead_code
)]
use proconio::input;
//use proconio::marker::Usize1;
// use proconio::marker::Chars;
// use std::collections::HashSet;
//use std::cmp::*;

const M: i64 = 1000_000_007i64;
fn main() {
    input! {
        Q:usize,
        lr:[(usize,usize);Q]

    }
    let mut rmax: usize = 200001;
    let mut primes = vec![true; rmax + 1];
    primes[0] = false;
    primes[1] = false;
    for i in 2..rmax + 1 {
        if !primes[i] {
            continue;
        }
        for j in 2..(rmax / i + 1) {
            primes[i * j] = false;
        }
    }

    let mut sum = vec![0usize; rmax + 1];
    //println!("{:?}", primes);
    // 1は素数ではない。
    sum[0] = 0;
    sum[1] = 0;
    for i in 1..rmax {
        sum[i] = sum[i - 1]
            + if i % 2 != 0 && primes[i] && primes[(i + 1) / 2] {
                1
            } else {
                0
            };
    }
    //println!("{:?}", sum);
    for (l, r) in lr {
        println!("{}", sum[r] - sum[l - 1]);
    }
}