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 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int q = sc.nextInt();
        boolean[] sieve = sieve(100000 + 10);
        int[] like = new int[100000 + 10 + 1];
        like[0] = 0;
        like[1] = 0;
        like[2] = 0;
        for (int i = 3; i < 100000 + 10; i += 2) {
            if (sieve[i] == true && sieve[(i + 1) / 2] == true) {
                like[i] = like[i-2] + 1;
            } else {
                like[i] = like[i-2];
            }
        }

        for (int i = 0; i < q; i++) {
            int l = sc.nextInt();
            int r = sc.nextInt();

            int ans = 0;
            if (l == 1) ans = like[r];
            else ans = like[r] - like[l - 2];
            System.out.println(ans);
        }
    }

    public static boolean[] sieve(int n) {
        final boolean[] sieve = new boolean[n+1];
        sieve[0] = false;
        sieve[1] = false;
        sieve[2] = true;
        for (int i = 3; i <= n; i += 2) {
            sieve[i] = true;
        }

        for (int i = 3; i * i <= n; i += 2) {
            if (sieve[i]) {
                for (int j = i * 2; j <= n; j += i) {
                    sieve[j] = false;
                }
            }
        }

        return sieve;
    }
}
