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)

#define _CRT_SECURE_NO_WARNINGS
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<functional>
#include<iomanip>
#include<iostream>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<utility>
#include<vector>

#define ll long long
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define FORR(i,a,b) for(int i=(a);i>=(b);--i)
#define REPR(i,n)   FORR(i,n,0)

using namespace std;

const int NMAX = 100000;
int isprime[NMAX + 1];


void eratos(int n) {
	REP(i, n) {
		isprime[i] = 1;
	}
	isprime[0] = 0;
	isprime[1] = 0;
	FOR(i, 2, n) {
		if (isprime[i]) {
			FOR(j, i + 1, n) {
				if (j%i == 0) {
					isprime[j] = 0;
				}
			}
		}
	}
}


int main() {
	int Q, l, r, S[100001] = { 0 };

	eratos(100001);
	FOR(i, 1, 100001) {
		if (isprime[i] && isprime[(i + 1) / 2]) {
			S[i] = S[i - 1] + 1;
		}
		else {
			S[i] = S[i - 1];
		}
	}

	cin >> Q;
	REP(i, Q) {
		cin >> l >> r;
		cout << S[r] - S[l - 1] << endl;
	}


	return 0;
}