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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;

class Main
{
	public static void main(String[] args) throws Exception
	{
		invoke(System.in);
	}

	public static void	invoke(InputStream is) throws IOException
	{
		// input
		BufferedReader	reader = new BufferedReader(new InputStreamReader(is));
		String	line = reader.readLine();

		boolean[]	isPrime = new boolean[50001];	// 3 > 1, 5 > 2, ...
		boolean[]	is2007 = new boolean[50001];	// 3 > 1, 5 > 2, ...

		int	maxN = 3;


		int	q = Integer.parseInt(line);
		for (int i = 0 ; i < q; ++i)
		{
			line = reader.readLine();
			String[]	va = line.split(" ");
			int l = Integer.parseInt(va[0]);
			int r = Integer.parseInt(va[1]);

			while (maxN <= r)
			{
				int	idx = maxN / 2;

				isPrime[idx] = isPrime(maxN);
				if (isPrime[idx])
				{
					int	nextN = (maxN + 1) / 2;
					if (isPrime[nextN / 2] && (nextN == 2 || nextN % 2 == 1))
					{
						is2007[idx] = true;
					}
				}

				maxN += 2;
			}

			int	result = 0;
			for (int j = l; j <= r; j += 2)
			{
				if (is2007[j / 2])
				{
					++result;
				}
			}
			System.out.println(result);
		}
	}

	private static boolean isPrime(int n)
	{
//		if (n % 2 == 0)
//		{
//			return	false;
//		}

		int	max = (int)(Math.sqrt(n));
		for (int i = 3; i <= max; i += 2)
		{
			if (n % i == 0)
			{
				return	false;
			}
		}
		return	true;
	}

}