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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class Main {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		String input = br.readLine();
		int q = Integer.parseInt(input);
		
//		System.out.println("start init");
		PrimeNumberGenerator pg = new PrimeNumberGenerator();
		
		int similar[] = new int[100000+1];
		for(int i = 1; i < similar.length ; i++){
			similar[i] = similar[i - 1];
			
			if(pg.isPrime(i) && pg.isPrime((i+1)/2)){
				similar[i] ++;
			}
		}
		
//		System.out.println("end init");
		
		for(int i = 0; i < q; i++){
			String[] tmpArray = br.readLine().split(" ");
			int l = Integer.parseInt(tmpArray[0]);
			int r = Integer.parseInt(tmpArray[1]);
			
			System.out.println(similar[r] - similar[l - 1]);
		}
	}

}

class PrimeNumberGenerator {
    private final int N = 100000;
    private boolean[] isPrime= new boolean[N + 1];
 
    public PrimeNumberGenerator() {
        Arrays.fill(isPrime, true);
 
        isPrime[0] = false;
        isPrime[1] = false;
 
        int limit = (int)Math.sqrt(N);
        for(int i = 2; i <= limit ; i++){
 
            if(isPrime[i] == false){
                continue;
            }
 
            for(int j = i * 2; j <= N; j += i){
                isPrime[j] = false;
            }
        }
    }
 
    public boolean isPrime(int index){
        return isPrime[index];
    }
}
