#include<bits/stdc++.h>
using namespace std;
int s[100001],q,prime[100001]={0},isprime[100001]={1,1,0},total=0,i,j,a,b,c=0;
int main()
{ 	
	for(i=2;i<=100000;i++)
	{
		if(isprime[i]==0)
		{
			prime[c++]=i;
			if(isprime[(i+1)/2]==0)
			{
				total++;
			}
		}
		s[i]=total;
		for(j=0;j<c&&i*prime[j]<=100000;j++)
		{
		    isprime[i*prime[j]]=1;
		    if(i%prime[j]==0)
		        break;
		}
	}
	scanf("%d",&q);
	for(i=0;i<q;i++)
	{
		scanf("%d %d",&a,&b);
		printf("%d\n",s[b]-s[a-1]);
	}
	return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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<bits/stdc++.h>
using namespace std;
int s[100001],q,prime[100001]={0},isprime[100001]={1,1,0},total=0,i,j,a,b,c=0;
int main()
{ 	
	for(i=2;i<=100000;i++)
	{
		if(isprime[i]==0)
		{
			prime[c++]=i;
			if(isprime[(i+1)/2]==0)
			{
				total++;
			}
		}
		s[i]=total;
		for(j=0;j<c&&i*prime[j]<=100000;j++)
		{
		    isprime[i*prime[j]]=1;
		    if(i%prime[j]==0)
		        break;
		}
	}
	scanf("%d",&q);
	for(i=0;i<q;i++)
	{
		scanf("%d %d",&a,&b);
		printf("%d\n",s[b]-s[a-1]);
	}
	return 0;
}
```