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.*;
//累積はプログラミングの世界での呼ばれ方
public class Main {
  public static void main(String[] args) {
    boolean[] isprime = new boolean[100010];
    for(int i = 0; i <= 100000 ; i++){
      isprime[i] = true;
    }

    isprime[0] = isprime[1] = false;

    for(int i = 2 ; i * i <= 100000 ; i++){
      int j = 2;
      while(i * j <= 100000){
        isprime[i*j] = false;
        j++;
      }
    }

    Scanner s = new Scanner(System.in);
    int Q = s.nextInt();
    int[] l1 = new int[Q] , r1 = new int[Q] , SR = new int[100010];
    for(int i = 0; i < Q; i++){
      //System.out.println(N + " " + Q + " " + S);
      l1[i] = s.nextInt();
      r1[i] = s.nextInt();
    }

    SR[0] = 0;
    for(int i = 1; i <= 100000; i++){
      if(isprime[i] && isprime[(i + 1) / 2])
        SR[i] = SR[i - 1] + 1;
      else
      SR[i] = SR[i - 1];
    }
    for(int i = 0; i < Q; i++){
      System.out.println(SR[ r1[i]] - SR[ l1[i] - 1]);
    }
  }
}
