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 OCaml (4.02.3) (judge verdict: Accepted)

let q = Scanf.scanf " %d" @@ (+) 0
let rmax = ref 0
let lrs = Array.init q @@ fun _ -> Scanf.scanf " %d %d" @@ fun a b ->
  rmax := max !rmax b; a, b
let is_prime n =
  if n <= 1 then false
  else
    let rec loop i =
      if i * i > n then true
      else if n mod i = 0 then false
      else loop @@ i + 1 in
    loop 2
let sums = Array.make (!rmax + 1) 0
let f n = if n mod 2 = 1 && is_prime n && is_prime @@ (n + 1) / 2 then 1 else 0
let _ =
  for i = 0 to !rmax - 1 do sums.(i + 1) <- sums.(i) + f (i + 1) done;
  let f (l, r) =
    Printf.printf "%d\n" @@ sums.(r) - sums.(l - 1)
  in
  Array.iter f lrs