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.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		execute22_1();
	}

	private static void execute22_1() {
		try (Scanner sc = new Scanner(System.in)) {
			int Q = sc.nextInt();
			int[][] q = new int[Q][2];

			for (int i = 0; i < Q; i++) {
				q[i][0] = sc.nextInt();
				q[i][1] = sc.nextInt();
			}

			boolean[] isSo = new boolean[100005];

			for (int i = 0; i < isSo.length; i++)
				isSo[i] = true;

			isSo[0] = false;
			isSo[1] = false;
			for (int i = 2; i*i < isSo.length; i++) {
				if(isSo[i]) for(int j = 2;i*j<isSo.length;j++) isSo[i*j] = false;
			}
			
			int[] sum = new int[100005];
			sum[0] = 0;
			for (int i = 1; i < sum.length; i++) {
				sum[i] = sum[i-1];
				if(isSo[i]&&isSo[(i+1)/2]) sum[i]++;
			}
				
			for (int i = 0; i < Q; i++) {
				int ans=sum[q[i][1]+1] - sum[q[i][0]-1];
				System.out.println(ans);
			}

		}
	}
}