Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define MAX 100000
#define TRUE 1
#define FALSE 0

int main(void){
  
  int i,j,n,q;
  int Is_Prime[MAX+1];
  int cs[MAX+1];
  
  Is_Prime[0] = FALSE;
  Is_Prime[1] = FALSE;
  for( i=2; i<=MAX; i++){
    Is_Prime[i] = TRUE;
  }
  
  for( i=2; i<=MAX/2; i++){
    if( Is_Prime[i] == TRUE ){
      //printf("%d is PRIME\n",i);
      for( j=i+1; j<=MAX; j++){
        if( j % i == 0){
          Is_Prime[j] = FALSE;
        }
      }
    }
  }
    
  for( i=1,cs[0]=0; i<=MAX; i++){
    if( Is_Prime[i] == TRUE && Is_Prime[(i+1)/2] ){
      cs[i] = cs[i-1] + 1;
    }
    else cs[i] = cs[i-1];
  }
  
  scanf("%d",&n);
  for( q=0; q<n; q++){
    scanf("%d %d",&i,&j);
    printf("%d\n",cs[j]-cs[i-1]);
  }
  
  return 0;
}