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 main()
{
	int q;
	cin >> q;
	bool prime[100001] = {};
	int n[100001] = {};
	prime[0] = true;
	prime[1] = true;
	for(int i = 2; i < 100001; i++)
	{
		for(int j = 2; j <= sqrt(i); j++)
		{
			if(!(i % j))
			{
				prime[i] = true;
				break;
			}
		}
	}
	for(int i = 3; i < 100000; i += 2)
	{
		n[i] = (!prime[i] && !prime[(i + 1) / 2]) + n[i - 2];
	}
	for(int i = 0, l, r; i < q; i++)
	{
		scanf("%d %d", &l, &r);
		printf("%d\n", n[r] - n[l - 2]);	
	}		
}