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
#include<bits/stdc++.h>
using namespace std;
int main(){
  vector<bool> p(100000,1);
  for(int i=2;i<p.size();i++){
    if(!p.at(i))
      continue;
    for(int j=2;i*j<p.size();j++)
      p.at(i*j)=0;
  }
  vector<bool> sp(100000,0);
  for(int i=3;i<p.size();i++)
    sp.at(i)=(p.at(i)&&p.at((i+1)/2));
  vector<int> c(100001,0);
  for(int i=3;i<p.size();i++)
    c.at(i+1)=c.at(i)+(sp.at(i)?1:0);
  int q;
  cin>>q;
  for(int _=0;_<q;_++){
    int l,r;
    cin>>l>>r;
    cout<<c.at(r+1)-c.at(l)<<endl;
  }
}