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.io.File;
import java.io.IOException;
import java.util.*;
 
public class Main {
 
	public static void main(String[] args) throws IOException {
		//File file = new File("input.txt");
		//Scanner in = new Scanner(file);
		
		Scanner in = new Scanner(System.in);
		
		int[] primeNumList = new int[100000];
		primeNumList[0] = -1;
		for(int i = 0; i < primeNumList.length; i++) {
			if(primeNumList[i] == -1) continue;
			for(int j = i + i + 1; j < primeNumList.length; j += (i + 1)) {
				primeNumList[j] = -1;
			}
		}
		
		/*
		for(int i = 0; i < 20; i++) {
			if(primeNumList[i] != -1) System.out.print((i + 1) + " ");
		}
		System.out.println();
		*/
		
		int[] susum = new int[50000];
		susum[0] = 0;
		susum[1] = 1;		
		for(int i = 2; i < susum.length; i++) {
			susum[i] = susum[i-1];
			if(primeNumList[2 * i] != -1 && primeNumList[i] != -1) susum[i]++;
		}
		
		/*
		for(int i = 0; i < 10; i++) {
			System.out.print(susum[i] + " ");
		}
		System.out.println();
		*/
		
		
		int Q = in.nextInt();
		
		for(int i = 0; i < Q; i++) {
			int L = in.nextInt();
			int R = in.nextInt();
			if(L == 1) System.out.println(susum[(R - 1) / 2]);
			else System.out.println(susum[(R - 1) / 2] - susum[((L - 1) / 2) - 1]);
		}
	}
}