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 n,l,r,sum[100001];
bool p[100001];
bool pd(int x)
{
	if (x==1) return false;
	for (int i=sqrt(x);i>=2;i--)
	  if (x%i==0) return false;
	return true;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=100000;i++)
	  p[i]=pd(i);
	for (int i=1;i<=100000;i++)
	{
		sum[i]=sum[i-1];
		if (p[i]==true&&p[(i+1)/2]==true) sum[i]++;
	}
	for (int i=1;i<=n;i++)
	{
		scanf("%d%d",&l,&r);
		printf("%d\n",sum[r]-sum[l-1]);
	}
	return 0;
}