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: Accepted)

#include<bits/stdc++.h>
#define MAX 1e5+1
using namespace std;

int main(){
	cin.tie(0),ios::sync_with_stdio(false);
	int q; cin>>q;
	vector<bool>sv(MAX,true); sv.at(0)=sv.at(1)=false;
	for(int i=2;i*i<MAX;++i){
		if(sv.at(i)){
			int j=i*i;
			while(j<MAX){
				sv.at(j)=false;
				j+=i;
			}
		}
	}
	vector<int>v(MAX,0);
	for(int i=3;i<MAX;++i){
		v.at(i)=v.at(i-1);
		if(sv.at(i)&&sv.at((i+1)/2))++v.at(i);
	}
	for(int i=0;i<q;++i){
		int l,r; cin>>l>>r;
		cout<<(v.at(r+1)-v.at(l-1))<<"\n"s;
	}
}