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>
#include <iostream>
#include<math.h>

using namespace std;



bool is_prime(int num)
{
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

bool is_nitakazu(int n) {
	if (is_prime(n)==true && is_prime((n + 1) / 2)==true) {
		return true;
	}
	else
	{
		return false;
	}
}


int main() {

	int n;
	int memo[100001];
	memo[1] = 0;
	memo[3] = 1;
	cin >> n;
	
	for (int i= 5; i < 100001; i+=2)
	{
		int temp;
		if (is_nitakazu(i) == true) {
			temp = 1;
		}
		else {
			temp = 0;
		}
		memo[i] =memo[i-2]+ temp;
	}

	int l, r;
	for (int i = 0; i < n; i++)
	{
		
		cin >> l >> r;
		if (l == 1) {
			l += 2;
		}
		cout << memo[r] - memo[l-2] << endl;
	}

	return 0;
}