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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import java.util.Scanner

case class Range(l: Int, r: Int) {
  val toList: scala.Range = l until r + 1
}


object Main {
  def read() = {
    val sc = new Scanner(System.in)
    val q = sc.nextInt()
    val ranges = for (_ <- 0 until q) yield Range(sc.nextInt(), sc.nextInt())
    (q, ranges)
  }

  def isPrime(n: Int): Boolean = {
    if (n <= 1) return false
    if (n == 2 || n == 3) return true
    val notPrime = (2 until math.sqrt(n).toInt + 1).exists(i => n % i == 0)
    return !notPrime
  }

  def like2017(n: Int): Boolean = isPrime(n) && isPrime((n + 1) / 2)

  def isOdd(n: Int) = n % 2 == 1

  def main(args: Array[String]): Unit = {
    val (q, ranges) = read()
    for (range <- ranges) {
      val result = range.toList.filter(isOdd).count(like2017)
      println(result)
    }
  }
}