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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
int prime[100005];
void sieve()
{
    prime[0]=0;prime[1]=0;
    for(int i=2;i<100005;i++) prime[i]=1;
    for(int i=2;i*i<100005;i++){
        //cout<<"POP"<<endl;
        if(prime[i]==1) {
          //  cout<<i<<endl;
            for(int j=i*i;j<=100005;j+=i) prime[j]=0;
        }
    }

}

int main()
{
    int n,m,c=0,t;
    sieve();
    cin>>t;
    while(t--){
    cin>>n>>m;
    for(int i=n;i<=m;i++){
    if(i%2!=0 && prime[i]==1){
       // cout<<i<<" "<<prime[i]<<endl;
        int h=(i+1)/2;
        if(prime[h]==1)  c++;

        }

    }
    cout<<c<<endl;
    c=0;
    }

}