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)

/*
  Lower bound: first element that is greater-or-equal.
  int i=lower_bound(a,a+n,x)-a;
  Upper bound: first element that is strictly greater.
  #include<boost/multiprecision/cpp_int.hpp>
  namespace mp=boost::multiprecision;
  mp::cpp_int u=1; for(unsigned i=1;i<=n;i++) u*=i;
*/
#pragma GCC optimize("O3")  
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/gcd(a,b)
#define mem(z,i) memset(z,i,sizeof(z))
#define eps 1e-7
#define I (int)
const int mod=(int)1e9+7;
const long long INF = 1000000000000000000LL;
const int WASTE=(int)2e5;
int power(int a,int b,int m=mod){
	if(b == 0) return 1;
	if(b == 1) return a;
	int x = power(a,b/2,m)%m;
	x = (x*x)%m;
	if(b%2) return (x*a)%m;
	return x;
}
//#define int long long
const int N=1e5+5;
bool prime[N];
void seive(){
	for(int x=2;x<N;x++){
		if(!prime[x]){
			for(int y=x*2;y<N;y+=x)
				prime[y]=1;
		}
	}
}
signed main() {
    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
    mem(prime,0);
    seive();
    int a[N];
    a[1]=0;
    for(int x=2;x<N;x++){
    	a[x]=a[x-1];
    	if(x%2){
    		if(!prime[x]&&!prime[(x+1)/2])
    			a[x]++;
    	}
    }
    int q;
    cin>>q;
    while(q--){
    	int x,y;
    	cin>>x>>y;
    	int ans=a[y]-a[x-1];
    	cout<<ans<<endl;
    }
    return 0;	
}