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 {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int Q = sc.nextInt();
    int[] l = new int[Q];
    int[] r = new int[Q];
    int max = 0;
    for (int i = 0; i < Q; i++) {
      l[i] = sc.nextInt();
      r[i] = sc.nextInt();
      max = Math.max(max, r[i]);
    }

    boolean[] prime = new boolean[max+5];
    for (int i = 2; i <= max; i++) {
      boolean isPrime = true;
      for (int j = 2; j*j <= i; j++) {
        if (i%j == 0) {
          isPrime = false;
          break;
        }
      }
      prime[i] = isPrime;
    }
    
    int[] cnt = new int[max+5];
    for (int i = 1; i <= max; i++) {
      if (prime[i] && prime[(i+1)/2]) {
        cnt[i] = cnt[i-1]+1;
      } else {
        cnt[i] = cnt[i-1];
      }
    }
    
    StringBuilder ans = new StringBuilder();
    for (int i = 0; i < Q; i++) {
      ans.append(cnt[r[i]] - cnt[l[i]-1]).append('\n');
    }

    System.out.print(ans.toString());
  }
}