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: Wrong Answer)

import java.util.*;
import java.util.ArrayList;

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];

        for(int i=0; i<Q; i++){
            l[i] = sc.nextInt();
            r[i] = sc.nextInt();
        }

        ArrayList<Integer> list  = new ArrayList<>();
        for(int i=2; i<=100000; i++){
            list.add(i);
        }

        for(int i=0; i<list.size(); i++){
            int x = i + 1;
            while(x<list.size()){
                if(list.get(x)%list.get(i)==0){
                    list.remove(x);
                }
                else{
                    x++;
                }
            }
        }
        int x = 0;
        int[] c = new int[100001];
        c[1] = 0;
        for(int i=2; i<=100000; i++){
            if(i==list.get(x)){
                if(list.contains((i+1)/2)){
                    c[i] = c[i - 1] + 1;
                }
                else{
                    c[i] = c[i - 1];
                }
                if(x!=list.size()-1){
                    x++;
                }
            }
            else{
                c[i] = c[i - 1];
            }
        }
        System.out.println(c[3] + " "+c[7]);
        for(int i=0; i<Q; i++){
            System.out.println(c[r[i]] - c[l[i] - 1]);
        }
    }
}