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 <cstdio>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int P_MAX=100000;
bool er[P_MAX+1];
void sieve(){
	rep(i,P_MAX+1) er[i]=(i>=2);
	for(int i=2;i*i<=P_MAX;i++) if(er[i]) for(int j=i*i;j<=P_MAX;j+=i) er[j]=false;
}

int main(){
	sieve();

	int ans[100001]={};
	for(int i=1;i<=100000;i++){
		ans[i]=ans[i-1];
		if(i%2==1 && er[i] && er[(i+1)/2]) ans[i]++;
	}

	int q; scanf("%d",&q);
	rep(i,q){
		int l,r; scanf("%d%d",&l,&r);
		printf("%d\n",ans[r]-ans[l-1]);
	}

	return 0;
}
