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.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	private static int[] prime = new int[100_000];
	static {
		int cnt = 0;
		prime[cnt++] = 2;
		prime[cnt++] = 3;
		for (int i = 5; i < prime.length; i += 2) {
			int j = 1;
			for (; j < cnt; j++) {
				if (i % prime[j] == 0) {
					break;
				}
			}
			if (j == cnt) {
				prime[cnt++] = i;
			}
		}
	}

	public static void main(String[] args) {

		try (Scanner sc = new Scanner(System.in); PrintWriter pw = new PrintWriter(System.out)) {
			int Q = Integer.parseInt(sc.nextLine());
			int[] l = new int[Q];
			int[] r = new int[Q];
			for (int i = 0; i < Q; i++) {
				l[i] = sc.nextInt();
				r[i] = sc.nextInt();
			}

			ArrayList<Integer> like2017 = new ArrayList<>();
			for (int i = 0; i < Q; i++) {
				int x = 0;
				for (int n = Integer.max(l[i], 3); n <= r[i]; n += 2) {
					if (like2017.contains(n)) {
						x++;
						continue;
					}
					if (isPrime(0, n) && isPrime(0, (n + 1) / 2)) {
						x++;
						like2017.add(n);
					}
				}
				pw.println(x);
			}
			pw.flush();
		}
	}

	private static boolean isPrime(int i, int n) {
		if (prime[i] < n) {
			return n % prime[i] == 0 ? false : isPrime(++i, n);
		}
		return true;
	}
}