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)

macro_rules! get {
    ($x:expr) => ($x.next().unwrap().parse().unwrap());
}
fn main() {
    use std::io::{Read, stdout, BufWriter, Write};
    let out = stdout();
    let mut out = BufWriter::new(out.lock());
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let mut sieve : [bool; 100001] = [true; 100001];
    sieve[0] = false;
    sieve[1] = false;
    for i in 2..100001 {
        if !sieve[i] {
            continue;
        }
        for j in 2..100001 {
            if j*i>100000 {
                break;
            }
            sieve[j*i] = false;
        }
    }
    let mut like2017 : [bool; 100001] = [false; 100001];
    for i in 2..100001 {
        if i%2==1&&sieve[i]&&sieve[(i+1)/2] {
            like2017[i]=true;
        }
    }
    let mut imos : [u32; 100001] = [0; 100001];
    for i in 1..100001 {
        if like2017[i] {
            imos[i] = imos[i-1]+1;
        } else {
            imos[i] = imos[i-1];
        }
    }
    for i in 0..100 {
        println!("imos[{}]={}", i, imos[i]);
    }
    let q : u32 = get!(iter);
    for _i in 0..q {
        let l : usize = get!(iter);
        let r : usize = get!(iter);
        writeln!(out, "{}", imos[r]-imos[l-1]).unwrap();
    }
}