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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;

// コンテスト中のみ
//#define int long long

#define ll long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define P pair<ll,ll>
#define ld long double
ll INF = (1LL << 60);
int MOD = 1000000007;

ll n;
ll Q;
bool isprime[200010];
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	n = 200000;
	rep(i, 200010)isprime[i] = true;
	isprime[0] = false;
	isprime[1] = false;
	for (int i = 2; i <= n; i++) {
		if (isprime[i]) {
			for (int j = 2 * i; j <= n; j += i)isprime[j] = false;
		}
	}
	vector<ll>sum(n + 2);
	rep(i, n + 1) {
		sum[i + 1] += sum[i];
		if (i%2==1&&isprime[i] && isprime[(i+1)/2]) {
			sum[i + 1]++;
		}
	}
	cin >> Q;
	rep(_, Q) {
		int l, r; cin >> l >> r;
		cout << sum[r + 1] - sum[l - 1] << endl;
	}
	return 0;
}