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<iostream>
#include<vector>
using namespace std;
#define MAXN 1e5+1

int Q,l,r;
vector<bool> isP(MAXN,true);
vector<int> cnt(MAXN,0);
int main(){
  cin>>Q;
  isP[0]=isP[1]=false;
  for(int i=2;i*i<=1e5;i++)if(isP[i]){
    for(int j=i*2;j<=1e5;j+=i){
      isP[j]=false;
    }
  }
  for(int i=3;i<=1e5;i+=2){
    cnt[i]=cnt[i-2];
    if(isP[i]&&isP[(i+1)/2]) cnt[i]++;
  }
  for(int i=0;i<Q;i++){
    cin>>l>>r;
    if(l==1) cout<<cnt[r]<<endl;
    else cout<<cnt[r]-cnt[l-2]<<endl;
  }
  return 0;
}