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<iostream>
const int N = 1e8+3;
bool death[N];
int primelist[N/15];
using namespace std;
int main() {
    int t, m, n, tail = 0;
    cin>>t;
    while(t--) {
      int sum = 0;
      cin>>m>>n;
      death[1] = 1;
      for(int i = 2;i <= n; ++i) {
      	if( !death[i] )
      		primelist[++tail] = i;
      	for(int j = 1; j <= tail; ++j) {
      		if(primelist[j]*i > n) break;
      		death[primelist[j]*i] = 1;
      		if(i%primelist[j] == 0) break;
      	}	
      }
      for(int i = m; i <= n; ++i) {
        if(i&1){
          if(!death[i]){
            if(!death[(i+1)/2]){
               sum++;
            }
          }
        }
      }
      cout<<sum<<"\n";
    }  
    return 0;
}