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.

Examples

Input

1
3 7


Output

2


Input

4
13 13
7 11
7 11
2017 2017


Output

1
0
0
1


Input

6
1 53
13 91
37 55
19 51
73 91
13 49


Output

4
4
1
1
1
2
#include<iostream>
#include<cmath>
using namespace std;
bool isp[100100],is2017[100100];
int Prefix[100100];
int main()
{
	int N;
	cin>>N;
	for(int i=0;i<100100;i++)
		isp[i]=true;
	isp[0]=isp[1]=false;
	for(int i=2;i<=sqrt(100100);i++)
		if(isp[i])
			for(int j=i*i;j<100100;j+=i)
				isp[j]=false;
	for(int i=1;i<100100;i+=2)
		is2017[i]=(isp[i]&&isp[(i+1)/2]);
	for(int i=1;i<100100;i++)
		Prefix[i]=Prefix[i-1]+is2017[i];
	while(N--)
	{
		int L,R;
		cin>>L>>R;
		cout<<Prefix[R]-Prefix[L-1]<<endl;
	}
	return 0;
}
