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>
using namespace std;
typedef long long int ll;
#define rep(i, n) for (int i = 0; i < (int)n; i++)

int main()
{
	int MAX_NUM = pow(10, 5);
	vector<int> sosuu;
	sosuu.push_back(2);
	sosuu.push_back(3);
	sosuu.push_back(5);
	sosuu.push_back(7);

	for (int i = 11; i <= MAX_NUM; i += 2) {
		bool isSosuu = true;
		for (int a : sosuu) {
			if (i % a == 0) {
				isSosuu = false;
				break;
			}
		}
		if (isSosuu) {
			sosuu.push_back(i);
		}
	}
	//outofbouds防止でテキトーにでかい数字をいれておく
	sosuu.push_back(10000007);

	//累積和をとる
	vector<int> countSum(MAX_NUM + 1);
	int sosuuIndex = 0;
	int likeNum = 0;
	for (int i = 0; i <= MAX_NUM; i++) {
		if (i >= sosuu[sosuuIndex]) {
			int v = sosuu[sosuuIndex];
			int v2 = (v + 1) / 2;
			if (binary_search(sosuu.begin(), sosuu.end(), v2)) {
				likeNum += 1;
			}
			sosuuIndex += 1;
		}
		countSum[i] = likeNum;
	}

	//問題解析
	int Q;
	cin >> Q;
	vector<int> Li(Q);
	vector<int> Ri(Q);
	rep(i, Q) {
		cin >> Li[i] >> Ri[i];
	}
	rep(i, Q) {
		int diff = countSum[Ri[i]] - countSum[Li[i] - 1];
		cout << diff << endl;
	}
	return 0;
}