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 scan = new Scanner(System.in);
    int q = scan.nextInt();
    long sum = 0;

    int[] a = new int[100001];

    for (int i=1;i<100001;i++){
      a[i] = find(i);
    }

    int count = 0;
    int[] b = new int[100001];
    for (int i=0;i<100001;i++){
      if (a[i] == 1){
        count += 1;
      }
      b[i] = count;
    }

    while (q > 0){
      sum = 0;
      int l = scan.nextInt();
      int r = scan.nextInt();
      sum = b[r] - b[l-1];
      System.out.println(sum);
      q--;
    }
  }

  static int find(int n){
    if (n == 3){
      return 1;
    }

    if (n%4 == 1){
      int a1 = (int) (n+1)/2;
      int a2 = 0;

      a2 += check(n);
      a2 += check(a1);

      if (a2 > 0){
        return 0;
      }
      return 1;


    }
    else{
      return 0;
    }
  }

  static int check(int n){
    int g = (int) Math.sqrt(n);

    if (g*g == n){
      return 1;
    }

    for (int i=g;i>=2;i--){
      if (n % i == 0){
        return 1;
      }
    }

    return 0;
  }

}
