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>
#include <string>
#include <vector>
#include <algorithm>
#include <string.h>

using namespace std;
typedef pair<int, int> P;
vector<int> prime;
bool isPrime[100001];

void primeCalc()
{
	memset(isPrime, 1, 100001);
	isPrime[0] = isPrime[1] = false;
	for (int i = 2; i < 100001; ++i)
	{
		if (isPrime[i])
		{
			prime.push_back(i);
			for (int j = i * 2; j < 100001; j += i)
			{
				isPrime[j] = false;
			}
		}
	}
}

void solve()
{
	primeCalc();
	int Q;
	cin >> Q;
	for (int i = 0; i < Q; ++i)
	{
		int l, r;
		cin >> l >> r;
		int cnt = 0;
		int to = *upper_bound(prime.begin(), prime.end(), r);
		for (vector<int>::iterator it = lower_bound(prime.begin(), prime.end(), l); *it != to; ++it)
		{
			if (isPrime[*it] && isPrime[(*it + 1) / 2])
			{
				++cnt;
			}
		}
		cout << cnt << endl;
	}
}

int main()
{
	solve();
	return(0);
}