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;
typedef long long ll;
typedef pair<int, int> pii;

int isPrime(int x)
{
	int i;
	if(x<2)return 0;
	else if(x==2)return 1;
	if(x%2==0)return 0;
	for(int i=3;i*i<=x;i+=2)
	{
		if(x%i==0)return 0;
	}
	return 1;
}

int main()
{
	int n,a,b,p[100005]={};
	for(int i=1;i<=100000;i+=2)
	{
		if(isPrime(i)&&isPrime((i+1)/2))p[i]++;
	}
	for(int i=1;i<=100000;i++)
	{
		p[i]+=p[i-1];
	}
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a>>b;
		cout<<p[b]-p[a-1]<<endl;
	}
	return 0;
}
	