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 n = sc.nextInt();
        for(int k = 0; k < n; k++) {
            boolean[] f = new boolean[100001];
            int[] c = new int[100002];
            int l = sc.nextInt();
            int r = sc.nextInt();
            for (int i = 2; i <= 100000; i++) {
                if (!f[i]) {
                    for (int j = i + i; j <= 100000; j += i) {
                        f[j] = true;
                    }
                }
            }
            for (int i = 3; i <= 100000; i += 2) {
                if (!f[i] && !f[(i + 1) / 2]) {
                    c[i]++;
                }
            }

            for (int i = 3; i <= 100000; i++) {
                c[i] += c[i - 1];
            }
            System.out.println( c[r] - c[l-1]);
        }

    }

    public static boolean isPrime(long a) {
        if(a < 2) return false;
        long i;
        for(i = 2; i * i <= a; i++) {
            if(a % i == 0) return false;
        }
        return true;
    }
}