#include<iostream>
using namespace std;
bool isp[1<<17];
int cnt[1<<17];
int main()
{
	for(int i=2;i<1<<17;i++)isp[i]=true;
	for(int i=2;i<1<<17;i++)
	{
		if(isp[i])for(int j=i+i;j<1<<17;j+=i)isp[j]=false;
	}
	for(int i=2;i<1<<17;i++)
	{
		cnt[i]=cnt[i-1];
		if(i%2==1&&isp[i]&&isp[(i+1)/2])cnt[i]++;
	}
	int Q;cin>>Q;
	for(;Q--;)
	{
		int l,r;cin>>l>>r;
		cout<<cnt[r]-cnt[l-1]<<endl;
	}
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include<iostream>
using namespace std;
bool isp[1<<17];
int cnt[1<<17];
int main()
{
	for(int i=2;i<1<<17;i++)isp[i]=true;
	for(int i=2;i<1<<17;i++)
	{
		if(isp[i])for(int j=i+i;j<1<<17;j+=i)isp[j]=false;
	}
	for(int i=2;i<1<<17;i++)
	{
		cnt[i]=cnt[i-1];
		if(i%2==1&&isp[i]&&isp[(i+1)/2])cnt[i]++;
	}
	int Q;cin>>Q;
	for(;Q--;)
	{
		int l,r;cin>>l>>r;
		cout<<cnt[r]-cnt[l-1]<<endl;
	}
}

```