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;
#define ran 100001
bool f[ran];int p[ran],s[ran];int lp;
int main(){
	f[0] = f[1] = true;
	for(int i=2; i<ran; i++){
		if(!f[i]){
			p[lp++] = i;
		}
		for(int j=0; j<lp; j++){
			int x = i*p[j];
			if(x>=ran)break;
			f[x] = true;
			if(i%p[j]==0)break;
		}
	}
	for(int i=3; i<ran; i++){
		s[i] = s[i-1];
		if(!f[i] && !f[(i+1)/2]){
			s[i]++;
		}
	}
	int q;
	for(scanf("%d",&q); q--;){
		int x,y;
		scanf("%d%d",&x,&y);
		printf("%d\n",s[y]-s[x-1]);
	}
	return 0;
}
