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>
#include <stdbool.h>

int main(void){
    bool sosu[100000];
    int ans[100000],q,l,r;
    ans[0]=0;
    for (int i=1; i<100000; i++){
        ans[i]=0;
        sosu[i]=true;
        for (int j=2; j<i; j++){
            if (i%j==0){
                sosu[i]=false;
                break;
            }
        }
        if (i%2!=0 && sosu[i]==true && sosu[(i+1)/2]==true){
            ans[i]=ans[i-1]+1;
        } else {
            ans[i]=ans[i-1];
        }
    }
    ans[0]=1;
    scanf("%d",&q);
    for (int i=0; i<q; i++){
        scanf("%d%d",&l,&r);
        printf("%d\n",ans[r]-ans[l-1]);
    }
	return 0 ;
}