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 {
	//java11

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int[] sum = new int[100001];
		sum[0] = 0;
		for(int i=1; i<=100000; i+=2) {
			if(like2017(i))
				sum[i] = sum[i-1] + 1;
			else
				sum[i] = sum[i-1];

			sum[i+1] = sum[i];
		}

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

			System.out.println(sum[r]-sum[l-1]);
		}

	}

	public static boolean like2017(int n) {
		if(n % 2 == 0)
			return false;

		if(isPrime(n) && isPrime((n+1)/2))
			return true;
		else
			return false;
	}

	//高速な素数判定
	public static boolean isPrime(int n) {
		if(n < 2)
			return false;

		for(int i=2; i<=Math.sqrt(n); i++) {
			if(n % i == 0)
				return false;
		}

		return true;
	}

}
