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 L 100010
int arr[L];
int res[L] = {0};
int main()
{
        for(int i = 0;i<L;i++){
                arr[i] = 1;
        }
        arr[0] = 0;
        arr[1] = 0;
        for(int i = 2;i<=L;i++){
                if(arr[i] == 1){
                        for(int j=i*2;j<=L;j+=i){
                                arr[j] = 0;
                        }
                }
        }
        for(int i = 3; i<L; i++){
                if(arr[i] == 1&&arr[(i+1)/2] == 1)
                        res[i] = 1;
                res[i] += res[i-1];
        }
        int T;
        scanf("%d",&T);
        while(T--){
                int a,b;
                scanf("%d %d",&a,&b);
                printf("%d\n",res[b] - res[a-1]);
        }
        return 0;
}