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;

bool isPrime[100001];
int np[100001];

int main(){
	for(int i=2;i<=100000;i++){
		isPrime[i]=true;
	}
	for(int i=2;i*i<=100000;i++){
		if(isPrime[i]){
			for(int j=i+i;j<100000;j=j+i){
				isPrime[j]=false;
			}
		}
	}
	np[0]=0;
	for(int i=1;i<=100000;i++){
		if(isPrime[i] and isPrime[(i+1)/2]){
			np[i]=np[i-1]+1;
		}else{
			np[i]=np[i-1];
		}
	}
	int q,l,r;
	cin>>q;
	while(q--){
		cin>>l>>r;
		cout<<np[r]-np[l-1]<<endl;
	}
}