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: Wrong Answer)

#include <stdio.h>
#include<math.h>
int count(int ll,int rr)
{
    int c=0;
    int max;
    int j=0;
    int ss;
    if(ll<=3&&rr>=3)c++;
    int lll;
    int llll=ll%4;
    if(llll==0||llll==1)lll=ll+1-llll;
    else lll=ll+5-llll;
    if(lll==1)lll+=4;
    for(int s=lll;s<rr+1;s+=4)
    {
            ss=(s+1)/2;
            max=sqrt(ss)+1;
            int u=3;
            while(u<=max&&j==0)
            {
                if(ss%u==0)j=1;
                u+=2;
            }
            max=sqrt(s)+1;
            int t=3;
            while(t<=max&&j==0)
            {
                if(s%t==0)j=1;
                t+=2;
            }
        if(j==0)c++;
        j=0;
    }
    return c;
}
 
 
int main(void)
{
    int q;
    int ans;
    scanf("%d",&q);
    int l[q];
    int r[q];
  	int unko08658932[100001];
  	unko08658932[0]=0;
  	for(int i=1;i<100001;i++)unko08658932[i]=count(i,i);
    for(int p=0;p<q;p++)scanf("%d%d",&l[p],&r[p]);
    for(int w=0;w<q;w++)
    {
        ans=unko08658932[r[w]]-unko08658932[l[w]-1];
        printf("%d\n",ans);
    }
}