Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#define ll int
#define mod 100000007
#define ln printf("\n");
#define mx 100001
using namespace std;
inline void in(ll &x){x=0;char c;bool f=0;while((c=getchar())>'9'||c<'0')if(c=='-')f=!f;x=c-48;while((c=getchar())<='9'&&c>='0')x=x*10+c-48;if(f)x=-x;}
inline void out(ll x){if(x<0){putchar('-');x=-x;}if(x>9)out(x/10);putchar(x%10+48);}
inline void outn(ll x){out(x);printf("\n");}
bool b[mx];
ll s[mx],q,l,r;
int main(){
	ll i,j;b[0]=b[1]=1;
	for(i=2;i<mx;i++)if(!b[i])for(j=2;j*i<mx;j++)b[i*j]=1;
	s[0]=s[1]=0;for(i=2;i<mx;i++){s[i]=s[i-1];if(!b[i]&&!b[(i+1)>>1])s[i]++;}
	in(q);
	while(q--){
		in(l);in(r);
		outn(s[r]-s[l-1]);
	}
}