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.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int q = sc.nextInt();
		int[] l = new int[q];
		int[] r = new int[q];
		int max = Integer.MIN_VALUE;
		for(int i = 0; i < q; i++) {
			int ltmp = sc.nextInt();
			int rtmp = sc.nextInt();
			max = Math.max(rtmp, max);
			l[i] = ltmp;
			r[i] = rtmp;
		}
		int[] sum = new int[max + 1];
		boolean[] pri = new boolean[max + 1];
		pri[2] = true;
		for(int i = 0; i < max + 1; i++) {
			sum[i] = 0;
		}
		for(int i = 3; i < max + 1; i = i + 2) {
			boolean flag = true;
			for(int j = 3; j < i; j = j + 2) {
				if(i % j == 0) {
					flag = false;
					break;
				}
			}
			if(flag)
				pri[i] = true;
			if(flag && pri[(i + 1) / 2]) {
				sum[i]++;
			}
			sum[i] += sum[i - 2];
		}
		for(int i = 0; i < q; i++) {
			int ans = 0;
			if(l[i] == 1) {
				ans += sum[r[i]] - sum[l[i]];
			} else {
				ans += sum[r[i]] - sum[l[i] - 2];
			}
			System.out.println(ans);
		}
	}

}