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 <iostream>
#include <cstdio>
#define N 100000
using namespace std;
int n, l, r, pr[N+5], v[N+5], s[N+5];
int main() {
	int i, j;
	for(i=2; i<N; i++) pr[i] = 1;
	for(i=2; i<N; i++) {
		if(pr[i]) {
			for(j=i*2; j<N; j+=i) pr[j]=0;
		}
	}
	for(i=1; i<N; i+=2) {
		if(pr[i] && pr[(i+1)/2]) v[i] = 1;
	}
	for(i=3; i<N; i+=2) {
		s[i] = s[i-2] + v[i];
	}
	scanf("%d", &n);
	while(n--) {
		scanf("%d%d", &l, &r);
		if(l==1) l=3;
		printf("%d\n", s[r] - s[l-2]);
	}
	return 0;
}
