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.

Examples

Input

1
3 7


Output

2


Input

4
13 13
7 11
7 11
2017 2017


Output

1
0
0
1


Input

6
1 53
13 91
37 55
19 51
73 91
13 49


Output

4
4
1
1
1
2
import java.util.*;

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

        int max = 100000;

        int[] c = new int[100001];
        c[0] = -1;
        c[1] = -1;

        for(int i = 2; i <= (int)Math.sqrt(max); i++){
            if(c[i] == 0){
                for(int j = i*i; j <= max; j = j+i){
                    c[j] = -1;
                }
            }
        }

        /* for(int i = 2; i <= max; i++){
            if(c[i] == 0)
                System.out.println(i);
        } */

        int[] s = new int[max+1];

        for(int i = 2; i <= max; i++){
            if(c[i] == 0 && c[(i+1)/2] == 0)
                s[i] = s[i-1]+1;
            else
                s[i] = s[i-1];
        }
        for(int i = 0; i < Q; i++){
            System.out.println(s[r[i]] - s[l[i]-1]);
        }
    }
}