#include <iostream>
#include <cstdio>
#define N 100005
using namespace std;
int n, l, r, pr[N], s[N];
int main() {
	int i, j;
	for(i=2; i<N; i++) if(!pr[i])
		for(j=i*2; j<N; j+=i) pr[j]=1;
	for(i=3; i<N; i+=2) {
		s[i] = s[i-2];
		if(!pr[i] && !pr[(i+1)/2]) s[i]++;
	}
	scanf("%d", &n);
	while(n--) {
		scanf("%d%d", &l, &r);
		printf("%d\n", s[r] - s[max(l-2, 1)]);
	}
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#include <cstdio>
#define N 100005
using namespace std;
int n, l, r, pr[N], s[N];
int main() {
	int i, j;
	for(i=2; i<N; i++) if(!pr[i])
		for(j=i*2; j<N; j+=i) pr[j]=1;
	for(i=3; i<N; i+=2) {
		s[i] = s[i-2];
		if(!pr[i] && !pr[(i+1)/2]) s[i]++;
	}
	scanf("%d", &n);
	while(n--) {
		scanf("%d%d", &l, &r);
		printf("%d\n", s[r] - s[max(l-2, 1)]);
	}
	return 0;
}
```