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: Runtime Error)

#include<iostream>
#include<string>
#include<algorithm>
#include<utility>
#include<vector>
#include<map>
#include<set>
#include<functional>
#include<iterator>
#include<math.h>

using namespace std;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	int q;
	cin >> q;

	int mem[5000] = {};
	mem[0]=-1;
	mem[1]=1;
	mem[2]=1;
	mem[3]=1;
	int l,r;
	int hl,hr;
	int ans=0;
	for(int i=0;i<q;i++){
		cin >> l >> r;

		if(l==1 || l==3){
			hl = 3;
			ans++; 		//hl = 2
		}
		else{
			hl = (l+1)/2;
			if(hl % 2 == 0) hl++;
		}
		hr = (r+1)/2;
											
		for(int j=hl;j<=hr;j+=2){ // jは3以上の奇数
			if(mem[j-1] == 0){  //初
				//cout << "first\n";
				int k;
				for(k=3; k <= j/2; k+=2){
					if(j%k == 0){
						mem[j-1] = -1;
						//cout << "a " <<  j << " " << k << mem[j-1] << "\n";
						break;
					}
				}
				if(k > j/2){

					int J = 2*j-1;
					int kk;
					for(kk=3; kk <= J/2; kk+=2){
						if(J%kk == 0){
							mem[j-1] = -1;
							//cout << "b " <<  j << " " << k << " " << kk << " " << mem[j-1] << "\n";
							break;
						}
					}
					if(kk > J/2){
						mem[j-1] = 1;
						//cout << "c " <<  j << " " << k << " " << kk << " " << mem[j-1] << "\n";
						ans++;
					}	
				}
			}else if(mem[j-1] == 1)	ans++;
		}
		cout << ans << "\n";
		ans = 0;
	}

	//cout << mem[1] << " " << mem[2] << " " << mem[3] << " " << mem[4] << "\n"; 
	return 0;
}