#include <iostream>
using namespace std;
int f[100001];
int c[100001];
int n, l, r;
int main()
{
	for (int i=2; i<=100000; i++) 
	{
    	if (!f[i])
    	{
    		for (int j=2*i; j<=100000; j+=i)
				f[j] = 1;
		}
  	}
	for (int i=3; i<=100000; i+=2)
	{
		if (!f[i] && !f[(i+1)/2])
    	c[i] += 1;
	}

	for (int i=3; i<=100000; i++)
	{
    	c[i] += c[i-1];
  	}

  	cin >> n;
  	while(n--) {
    	cin >> l >> r;
    	cout << c[r]-c[l-1] << endl;
  	}
  	return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <iostream>
using namespace std;
int f[100001];
int c[100001];
int n, l, r;
int main()
{
	for (int i=2; i<=100000; i++) 
	{
    	if (!f[i])
    	{
    		for (int j=2*i; j<=100000; j+=i)
				f[j] = 1;
		}
  	}
	for (int i=3; i<=100000; i+=2)
	{
		if (!f[i] && !f[(i+1)/2])
    	c[i] += 1;
	}

	for (int i=3; i<=100000; i++)
	{
    	c[i] += c[i-1];
  	}

  	cin >> n;
  	while(n--) {
    	cin >> l >> r;
    	cout << c[r]-c[l-1] << endl;
  	}
  	return 0;
}

```