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;

const int N = 1e5 + 100;

int q, l, r, cnt[N];
bool mark[N];

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	for (int i = 2; i < N; i++)
		if(mark[i] == false) {
			for (int j = i << 1; j < N; j += i)
				mark[j] = true;
		}
	for (int i = 3; i < N; i++)
		cnt[i] = cnt[i - 1] + (mark[i] == false && mark[(i + 1) >> 1] == false);
	cin >> q;
	while(q--) {
		cin >> l >> r;
		cout << cnt[r] - cnt[l - 1] << '\n';
	}
	return 0;
}
