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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        static final int SIZE = (int) 1e5 + 100;
        static boolean[] isPrime = new boolean[SIZE];
        static int[] similarCount = new int[SIZE];

        static {
            Arrays.fill(isPrime, true);
            isPrime[1] = false;
            for (int i = 2; i * i < SIZE; i++) {
                if (!isPrime[i]) continue;
                for (int j = 2, mul = j * i; mul < SIZE; j++, mul = j * i) {
                    isPrime[mul] = false;
                }
            }
            similarCount[0] = similarCount[1] = 0;
            for (int i = 3; i < SIZE; i += 2) {
                if (isPrime[i] && isPrime[(i + 1) / 2]) similarCount[i] = similarCount[i - 2] + 1;
                else similarCount[i] = similarCount[i - 2];
            }
        }

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int q = in.nextInt();
            for (int i = 0; i < q; i++) {
                int l = in.nextInt(), r = in.nextInt();
                out.println(similarCount[r] - similarCount[Math.max(0, l - 2)]);
            }
        }

    }
}

