#include <bits/stdc++.h>
using namespace std;

int main(){
  vector<bool> flag(100001,false);
  vector<int> num(100001,0);
  flag[2]=true;
  for(int i=3;i<100001;i+=2){
    bool f=true;
    for(int j=3;j*j<=i;j++){
      if(i%j==0){
        f=false;
        break;
      }
    }if(f) flag[i]=true;
  }for(int i=3;i<100001;i++){
    if(flag[i] && flag[(i+1)/2]) num[i]++;
    num[i]+=num[i-1];
  }int q; cin>>q;
  for(int i=0;i<q;i++){
    int l,r; cin>>l>>r;
    cout<<num[r]-num[l-1]<<endl;
  }
}
### Prompt

Your task is to create 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 main(){
  vector<bool> flag(100001,false);
  vector<int> num(100001,0);
  flag[2]=true;
  for(int i=3;i<100001;i+=2){
    bool f=true;
    for(int j=3;j*j<=i;j++){
      if(i%j==0){
        f=false;
        break;
      }
    }if(f) flag[i]=true;
  }for(int i=3;i<100001;i++){
    if(flag[i] && flag[(i+1)/2]) num[i]++;
    num[i]+=num[i-1];
  }int q; cin>>q;
  for(int i=0;i<q;i++){
    int l,r; cin>>l>>r;
    cout<<num[r]-num[l-1]<<endl;
  }
}
```