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
 *
 * @author ZYCSwing
 */
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 {
        private static final int N = 100010;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            boolean[] prime = new boolean[N];
            Arrays.fill(prime, true);
            prime[0] = prime[1] = false;
            for (int i = 2; i < N; ++i) {
                if (prime[i]) {
                    for (int j = 2 * i; j < N; j += i) {
                        prime[j] = false;
                    }
                }
            }

            int[] cnt = new int[N];
            Arrays.fill(cnt, 0);

            for (int i = 2; i < N; ++i) {
                cnt[i] = cnt[i - 1];
                if (i % 2 == 1 && prime[i] && prime[(i + 1) / 2]) {
                    ++cnt[i];
                }
            }

            int q = in.nextInt();
            int l, r;
            for (int i = 0; i < q; ++i) {
                l = in.nextInt();
                r = in.nextInt();
                out.println(cnt[r] - cnt[l - 1]);
            }
        }

    }
}

