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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
class Program
{
	static void Main()
	{
  bool[] isPrimes = makePrimeList(100000);//i-1が素数かどうか
  int[] trueCount = new int[100000];//i-1以下の条件を満たす数の個数

    int nowTrueCount = 0;
    for(int i=0; i<trueCount.Length; i+=2)//1以上i以下の素数の個数
    {
      if(isPrimes[i] && isPrimes[(i+1)/2]) nowTrueCount++;
      trueCount[i] = nowTrueCount;
    }
  
    int q = int.Parse(Console.ReadLine());
    for(int i = 0; i < q; i++)//q個のクエリに答える
    {
      int[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
      Console.WriteLine(trueCount[nums[1]-1] - trueCount[nums[0]-1]);
    }

	}

  static bool[] makePrimeList(int maxPrime)//引数max以下の素数のリストを作成。エラトステネスの篩を用いる。
  {
    bool[] isPrimes = new bool[maxPrime];//素数リスト
    isPrimes[0] = false;
    for(int i = 1; i < maxPrime; i++) isPrimes[i] = true;
    
    for(int i = 2; i < Math.Sqrt(maxPrime); i++)
    {
      if(isPrimes[i])
      {
        for (int j = i*2; j <= maxPrime; j+=i)
        {
          isPrimes[j-1] = false;
        }
      }
    }

    return isPrimes;
  }
  
}