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<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int b[1000000+5],q,l,r;
bool check(int x){
	bool flag=true;
	if(x==1){return false;}
	for(int i=2;i<=sqrt(x);i++){
		if(x%i==0){
			flag=false;
			break;
		}
	}
	return flag;
}
int main(){
	for(int i=1;i<=100000;i++){
		b[i]=b[i-1];
		if(i%2==1&&check(i)&&check((i+1)/2)){
			b[i]++;
		}
	}
	scanf("%d",&q);
	for(int i=1;i<=q;i++){
		scanf("%d%d",&l,&r);
		if(l>r){swap(l,r);}
		printf("%d\n",b[r]-b[l-1]);
	}
	return 0;
}