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;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Iterator;
import java.util.Collections;
import java.util.Comparator;
import java.util.Random;
import java.util.Map;
import java.util.HashMap;
 
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 maxR=0;
                int minL=10000;
                for(int i=0; i<Q; i++){
                    l[i] = sc.nextInt();
                    r[i] = sc.nextInt();
                    maxR = Math.max(maxR,r[i]);
                    minL = Math.min(minL,l[i]);
                }
                ArrayList<Integer> sosu = new ArrayList<Integer>();
                //ArrayList<Integer> niteru = new ArrayList<Integer>();
                //素数を(minL+1)/2からmaxRまで列挙
                int[] ans = new int[Q];
                for(int i=0; i<Q; i++)ans[i]=0;
                boolean ch = true;
                int start = Math.max((minL+1)/2, 3);
                if(start % 2 ==0)start++;
                sosu.add(2);
                
                for(int i=start; i<=maxR; i+=2){
                    ch = true;
                    for(int j=2; j<(i/2)+1; j++){
                        if(i%j == 0){
                            ch =false;
                            break;
                        }
                    }
                    if(ch){
                    sosu.add(i);
                    if(sosu.indexOf((i+1)/2) != -1){
                        //niteru.add(i);
                        for(int k=0; k<Q; k++){
                            if(l[k]<=i && i<=r[k]){
                                ans[k] ++;
                            }
                        }
                    }
                    }
                }
                
                for(int i=0; i<Q; i++){
                    System.out.println(ans[i]);
                }
        }
}