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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Program
{
    static void Main(string[] args)
    {
        var num = int.Parse(Console.ReadLine());

        var primeFlag = new bool[100000];
        for (int i = 0; i < 100000; i++)
        {
            primeFlag[i] = IsPrime(i);
        }

        for (int i = 0; i < num; i++)
        {
            var count = 0;
            var str = Console.ReadLine();
            int min = int.Parse(str.Split(' ')[0]);
            int max = int.Parse(str.Split(' ')[1]);
            for(int n = min; n<=max; n+=2)
            {
                if (primeFlag[n] && primeFlag[(n + 1) / 2])
                {
                    count++;
                }
            }
            Console.WriteLine(count);
        }
    }

    public static bool IsPrime(int num)
    {
        if (num < 2) return false;
        else if (num == 2) return true;
        else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

        double sqrtNum = Math.Sqrt(num);
        for (int i = 3; i <= sqrtNum; i += 2)
        {
            if (num % i == 0)
            {
                // 素数ではない
                return false;
            }
        }

        // 素数である
        return true;
    }
}