#include <bits/stdc++.h>
using namespace std;
int sum[100007],q,l,r;
int check[100007];
int main(){
	ios_base::sync_with_stdio(0);
	check[0]=check[1]=1;
	sum[0]=sum[1]=1;
	for(int i=2;i<=100000;i++){
		sum[i]=sum[i-1]+(check[i]==0 && check[(i+1)/2]==0);
		if(check[i]) continue;
		for(int j=2*i;j<=100000;j+=i){
			check[j]=1;
		}
	}
	cin>>q;
	while(q--){
		cin>>l>>r;
		cout<<sum[r]-sum[l-1]<<"\n";
	}
}

### Prompt

In cpp, your task is to solve 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 sum[100007],q,l,r;
int check[100007];
int main(){
	ios_base::sync_with_stdio(0);
	check[0]=check[1]=1;
	sum[0]=sum[1]=1;
	for(int i=2;i<=100000;i++){
		sum[i]=sum[i-1]+(check[i]==0 && check[(i+1)/2]==0);
		if(check[i]) continue;
		for(int j=2*i;j<=100000;j+=i){
			check[j]=1;
		}
	}
	cin>>q;
	while(q--){
		cin>>l>>r;
		cout<<sum[r]-sum[l-1]<<"\n";
	}
}

```