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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long ll;
typedef pair<ll,ll> P;
const ll mod=1000000007;
const ll LINF=1LL<<60;
const int INF=1<<30;

vector<ll> prime(10000);
bool is_prime[10000];

ll sieve(ll n){
	ll p = 0;
	for(int i = 0; i <= n; i++) {//初期化
		is_prime[i] = true;
	}
	is_prime[0] = is_prime[1] = false;
	for(int i = 2; i <= n; i++) {
		if(is_prime[i]){
			prime[p++] = i;
			for(int j = 2 * i; j <= n; j+= i) {
				is_prime[j] = false;
			}
		}
	}
	return p;
}

int main(){
	int q;cin>>q;
	vector<P> v(q);
	for(int i = 0; i < q; i++) {
		ll l,r;cin>>l>>r;
		v[i] = mp(l,r);
	}

	sieve(2020);
	vector<int> a(2020,0);
	for(int i = 3; i <= 2017; i+= 2) {
		if((is_prime[i]) && (is_prime[(i + 1) / 2])){
			a[i]= 1 + a[i - 2];
		}
		else{
			a[i] = a[i - 2];
		}
	}
	for(int i = 0; i < q; i++) {
		cout << a[v[i].sc] - a[max((v[i].fs - 2LL),1LL)] << endl;
	}
	return 0;
}