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.Scanner;
public class Main{
	
	static int p = 100000;
	
	static boolean prime(int num) {
		if (num < 2) return false;
	    else if (num == 2) return true;
	    else if (num % 2 == 0) return false; 

	    double sqrtNum = Math.sqrt(num);
	    for (int i = 3; i <= sqrtNum; i += 2)
	    {
	        if (num % i == 0)
	        {
	            return false;
	        }
	    }
	    return true;

	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int q = sc.nextInt();
		
		int []a = new int [p+1];
		
		for(int i = 0; i < p ; i++) {
			if(i % 2 == 0) {
				a[i+1] += a[i];
				continue;
			}
			if(prime(i)&&prime((i+1)/2)) {
				a[i]++;
			}
			a[i+1] += a[i];
		}
		int []ans = new int [q];
		for(int j = 0; j < q ; j++) {
			int l = sc.nextInt();
			int r = sc.nextInt();
			
			ans[j] = a[r] - a[l-1];
		}
		
		for(int i = 0 ; i < q ;i++) {
			System.out.println(ans[i]+" ");
		}
	

	}

}
