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.*;

public class Main {
        static int RMAX = 100001;
        static boolean[] primes = new boolean[RMAX];
        static int[] count2017num = new int[RMAX];
        public static void main(String args[]) {
                Scanner sc = new Scanner(System.in);
                StringBuilder sb = new StringBuilder(1000000);
                int Q = sc.nextInt();
                e();
                for (int i = 0; i < Q; i++) {
                        int l = sc.nextInt();
                        int r = sc.nextInt();
                        sb.append(count2017num[r] - count2017num[l - 1]);
                        sb.append('\n');
                }
                System.out.print(sb);
        }

        static void e() {
                for (int i = 2; i < RMAX; i++) {
                        primes[i] = true;
                }
                for (int i = 2; i < RMAX; i++) {
                        if (!primes[i]) {
                                continue;
                        }
                        for (int j = 2; j * i < RMAX; j++) {
                                primes[i * j] = false;
                        }
                }
                int counter = 0;
                for (int i = 1; i < RMAX; i++) {
                        if (primes[i] && primes[(i + 1) / 2]) counter++;
                        count2017num[i] = counter;
                }
        }
}
