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;

vector<bool> isPrime(100001, true);
vector<int> like(100001);

int main() {
	isPrime[0]=false;
	isPrime[1]=false;
	for (int i=2; i<100001; i++) {
		if (isPrime[i])
			for (int j=2; i*j<100001; j++)
				isPrime[i*j]=false;
	}

	for (int i=1; i<100001; i++) {
		if (isPrime[i-1] and isPrime[(i)/2]) like[i]=like[i-1]+1;
		else like[i]=like[i-1];
	}
	
	int q; cin>>q;
	while (q--) {
		int l,r; cin>>l>>r;
		cout << like[r+1]-like[l] << endl;
	}
	return 0;
}