#include <iostream>
using namespace std;
int a[100001],s[100001],i,i2,n,l,r;
int main()
{
	a[0]=a[1]=1;
	for(i=2;i<=100000;i++)
	{
		if(a[i]==0)
		{
			for(i2=i*2;i2<=100000;i2+=i)
				a[i2]=1;
		}
	}
	for(i=1;i<=100000;i++)
	{
		s[i]=s[i-1];
		if(a[i]==0&&a[(i+1)/2]==0)s[i]++;
	}
	cin>>n;
	while(n--)
	{
		cin>>l>>r,cout << s[r]-s[l-1]<<'\n';
	}
}
### 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;
int a[100001],s[100001],i,i2,n,l,r;
int main()
{
	a[0]=a[1]=1;
	for(i=2;i<=100000;i++)
	{
		if(a[i]==0)
		{
			for(i2=i*2;i2<=100000;i2+=i)
				a[i2]=1;
		}
	}
	for(i=1;i<=100000;i++)
	{
		s[i]=s[i-1];
		if(a[i]==0&&a[(i+1)/2]==0)s[i]++;
	}
	cin>>n;
	while(n--)
	{
		cin>>l>>r,cout << s[r]-s[l-1]<<'\n';
	}
}
```