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.*;
 
class Main
{
    public static int[] prime = new int[100001];
    public static int[] like = new int[100001];
    public static int index1 = 0;
    public static int index2 = 0;
 
    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();
        }
 
        Prime();
        LikeMake();
 
        for(int i = 0; i < Q; i++){
            int L_index = Arrays.binarySearch(like, L[i]);
 
            if(L_index < 0){
                L_index = (-1) * L_index - 1; 
            }
 
            int R_index = Arrays.binarySearch(like, R[i]);
 
            if(R_index < 0){
                R_index = (-1) * R_index  - 2;
            }
            System.out.println( R_index - L_index + 1);
        }
    }
    static void Prime() // 素数生成
    {
        boolean bl = true;
    
        prime[0] = 2;
 
        for(int i = 1; i < prime.length; i++){
            prime[i] = 10000007;
        }
 
        for(int num = 3; num < prime.length; num += 2){
            bl = true;;
            for(int j = 0; Math.pow(prime[j], 2) <= num; j++){
                if( num % prime[j] == 0){
                    bl = false;
                    break;
                }
            }
            if(bl == true){
                prime[++index1] = num;
            }
        }
    }
    static boolean LikeJudge(int p)
    {
        boolean ans = false;            
        if( Arrays.binarySearch(prime, (p + 1) / 2) >= 0){
            ans = true;
        }
        return ans;
    }
    
    static void LikeMake()
    {
        for(int i = 0; i < like.length; i++){
            like[i] = 100001;
        }
        for(int i = 1; i < prime.length; i++){
            if(LikeJudge(prime[i]) == true){
                like[index2] = prime[i];
                index2++;
            }
        }
    }
}