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;

public class Main {

	public static void main(String[] args) {
		try(BufferedReader bf = new BufferedReader(new InputStreamReader(System.in))){
			int max = 100010;
			//探索用配列
			boolean[] target = new boolean[max];
			Arrays.fill(target, true);
			target[0] = false;
			target[1] = false;

			int sqrt = (int)Math.sqrt(target.length);
			for(int i=2; i<=sqrt; i++) {
				int firstNum = i;
				if (target[i]) {
					for (int j=i*firstNum; j<target.length; j+=firstNum) {
						target[j] = false;
					}
				}
			}
			int[] a = new int[max];
			for (int i=2; i<target.length; i++) {
				if(i%2==0) continue;
				if (target[i] && target[(i+1)/2]) {
					a[i] = 1;
				}
			}
			//累積和
			int[] sum = new int[max+1];
			for(int i = 0;i < max;i++) {
				sum[i+1] = sum[i] + a[i];
			}
			//Query処理
			int Q = Integer.parseInt(bf.readLine());
			for(int q = 0;q < Q;q++) {
				String[] s = bf.readLine().split(" ");
				int l = Integer.parseInt(s[0]);
				int r = Integer.parseInt(s[1])+1;
				System.out.println(sum[r] - sum[l]);
			}
		}catch(IOException ioe) {
			ioe.getMessage();
		}
	}
}