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 <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <math.h>

#define MAX 10e5

using namespace std;

bool IsPrime(int num) {
	if (num < 2) return false;
	else if (num == 2)return true;
	else if (num % 2 == 0) return false;

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2) {
		if (num % i == 0) {
			return false;
		}
	}
	return true;
}

int main() {
	int q;
	cin >> q;
	vector<int> l;
	vector<int> r;
	for (int i = 0; i < q; i++) {
		int a, b;
		cin >> a >> b;
		l.push_back(a);
		r.push_back(b);
	}

	vector<int> sum(MAX,0);
	vector<int> isPrime(MAX,0);

	for (int i = 3; i < MAX; i += 2) {
		if (IsPrime(i))
			isPrime[i] = 1;
	}

	for (int i = 2; i < MAX / 2; i++) {
		if (isPrime[2 * i - 1] != 1)
			continue;
		if (IsPrime(i)) {
			//cout << 2 * i - 1 << "は素数" << endl;
			continue;
		}
		else {
			isPrime[2 * i - 1] = 0;
		}
	}

	sum[3] = 1;

	for (int i = 5; i < MAX; i += 2) {
		sum[i] = sum[i - 2] + isPrime[i];
	}

	for (int i = 0; i < q; i++) {
		int x = sum[r[i]] - sum[l[i]];
		if (isPrime[l[i]])
			x += 1;
		cout << x << endl;
		//cout << r[i] << " " << sum[r[i]] << endl << l[i] << " " << sum[l[i]] << endl;
	}

	for (int i = 0; i < q; i++) {
		//cout << sum[i] << endl;
	}


}