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: Time Limit Exceeded)

#include <stdio.h>
int jud(int n);
int main()
{
        int T;
        scanf("%d",&T);
        while(T--){
                int a,b,i,cnt = 0;
                scanf("%d %d",&a,&b);
                for(i = a; i<=b; i++){
                        if(jud(i)){
                                cnt++;
                        }
                }
                printf("%d",cnt);
                if(T!=0)
                        printf("\n");
        }
        return 0;
}

int jud(int n){
        if(n==1||n==2) return 0;
        int flag = 1;
        int i;
        for(i = 2; i<n; i++){
                if(n%i==0) flag=0;
        }
        int n2 = (n+1)/2;
        for(i = 2; i<n2; i++){
                if(n2%i==0) flag=0;
        }
        return flag;
}