#include<bits/stdc++.h>
using namespace std;
bool f[100001];
int C[100002];
int N, L, R;
int main() {
  cin>>N;
	for (int i = 2; i <= 100000; i++)if (!f[i])for (int j = i + i; j <= 100000; j += i) f[j] = true;
	for (int i = 3; i <= 100000; i+=2) {
		if (!f[i] && !f[(i + 1) / 2]) C[i]++;
	}
	for (int i = 3; i <= 100000; i++) C[i] += C[i - 1];
	while (N--) {
		cin >> L >> R;
		cout << C[R] - C[L - 1] << endl;
	}
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
bool f[100001];
int C[100002];
int N, L, R;
int main() {
  cin>>N;
	for (int i = 2; i <= 100000; i++)if (!f[i])for (int j = i + i; j <= 100000; j += i) f[j] = true;
	for (int i = 3; i <= 100000; i+=2) {
		if (!f[i] && !f[(i + 1) / 2]) C[i]++;
	}
	for (int i = 3; i <= 100000; i++) C[i] += C[i - 1];
	while (N--) {
		cin >> L >> R;
		cout << C[R] - C[L - 1] << endl;
	}
}
```