#include<bits/stdc++.h>
using namespace std;
int main(){
    vector<int> a(100001);
    for(int i=2;i<=100000;i++){
        for(int j=i*2;j<=100000;j+=i)a[j]++;
    }
    vector<int> cnt(100001);
    for(int i=1;i<=100000;i++){
        if(i!= 1 and i%2 and a[i]==0 and a[(i+1)/2]==0)cnt[i]=1;
        cnt[i]+=cnt[i-1];
    }
    int q;scanf("%d",&q);
    while(q--){
        int l,r;scanf("%d%d",&l,&r);
        printf("%d\n",cnt[r]-cnt[l-1]);
    }
}
### 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 main(){
    vector<int> a(100001);
    for(int i=2;i<=100000;i++){
        for(int j=i*2;j<=100000;j+=i)a[j]++;
    }
    vector<int> cnt(100001);
    for(int i=1;i<=100000;i++){
        if(i!= 1 and i%2 and a[i]==0 and a[(i+1)/2]==0)cnt[i]=1;
        cnt[i]+=cnt[i-1];
    }
    int q;scanf("%d",&q);
    while(q--){
        int l,r;scanf("%d%d",&l,&r);
        printf("%d\n",cnt[r]-cnt[l-1]);
    }
}
```