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>

int main()
{
    int judge(int);
	int a[100100];
	int i;
	int k=0;
	for(i=3;i<=110000;i+=2)
    {
        if(judge(i)&&judge((i+1)/2))
        {
            a[k]=i;
            k++;
        }
    }

	int t,m,n,j;
	scanf("%d",&t);
	while(t--)
    {
        int sum=0;
        scanf("%d%d",&m,&n);
        for(j=0;a[j]<=n;j++)
        {
            if(a[j]>=m)
            {
                sum++;
            }
        }
        printf("%d\n",sum);
    }
	return 0;
}
int judge(int x)
{
    int i;
    for(i=2;i*i<=x;i++)
        if(x%i==0) return 0;
    return 1;
}