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 ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define REP(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) std::cout << x << '\n'
#define all(a) a.begin(), a.end()
#define pb push_back
#define sz(x) ((int)(x).size())
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g * b; }
const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INF = (int)1e9;
const ll  inf = (ll)1e18;
const int mod = 1000000007;
const int MAX_N = 100010;
const double PI = acos(-1);

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(10);
	/*------------------------------*/
	
	//eratos
	const int MAX = 100010;
	vector<int> is_prime(MAX, 1);
	is_prime[0] = 0, is_prime[1] = 0;
	for (int i = 2; i < MAX; i++) {
		if (!is_prime[i]) continue;
		for (int j = i * 2; j < MAX; j += i) {
			is_prime[j] = 0;
		}
	}

	//2017-like
	vector<int> a(MAX, 0);
	for (int i = 0; i < MAX; i++) {
		if (i % 2 == 0) continue;
		if (is_prime[i] && is_prime[(i + 1) / 2]) {
			a[i] = 1;
		}
	}

	//累積和
	vector<int> s(MAX + 1, 0);
	for (int i = 0; i < MAX; i++) {
		s[i + 1] = s[i] + a[i];
	}

	int q;
	cin >> q;
	rep(i, q) {
		int l, r;
		cin >> l >> r;
		cout << s[r + 1] - s[l] << endl;
	}

	return 0;
}