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: Wrong Answer)


#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;


int ans;
inline bool is_prime(int n)
{
	if (n <= 2)
	{
		return true;
	}
	else if(n%2!=0)
	{
		for (int i = 2; i*i <= n; i++)
		{
			if (n%i == 0&&((n+1)/2)%i==0)
				return false;
		}
		return true;
	}
	return false;
}

int main()
{
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	while(n--)
	{
		int l, r;
		cin >> l >> r;
		for (; l <= r; l++)
		{
			if (is_prime(l))
			{
				ans++;
			}
		}
		cout << ans << endl;
		ans = 0;
	}
    return 0;
}