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 {

	boolean isprime[] = new boolean[100001];
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		Main main =new Main();
		main.sieve();
		int primeCount[]=new int[100001];
		int count =0;
		for(int i=0;i<=100000;i++) {
			if(main.isprime[i]==true&&main.isprime[(i+1)/2]==true) {
				count ++;
			}
			primeCount[i] =count;
		}
		int Q= sc.nextInt();
		for (int i=0;i<Q;i++) {
			int l= sc.nextInt();
			int r =sc.nextInt();
			if (l==r) {
				if (main.isprime[l]==true&&main.isprime[(l+1)/2]==true) {
					System.out.println(1);
				} else {
					System.out.println(0);
				}
			} else {
				int c =primeCount[r]-primeCount[l];
				if (main.isprime[l]==true&&main.isprime[(l+1)/2]==true) {
					c++;
				}
				System.out.println(c);
			}
		}
	}

	public void  sieve() {
		for (int i=0;i<=100000;i++) {
			isprime[i]=true;
		}
		isprime[0] =false;
		isprime[1] =false;
		for (int i=2;i<=100000;i++) {
			if (isprime[i]==true) {
				for(int j=2*i;j<=100000;j+=i) {
					isprime[j] =false;
				}
			}
		}
	}
}