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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mod 1000000007

bool f [100001];
int c [100002];
int N ,L , R ;
int main ()
{
for ( int i =2; i <=100000; i ++) if (! f [ i ]) for ( int j = i + i ;j <=100000; j += i )
f[ j ]= true ;
for ( int i =3; i <=100000; i +=2) if (! f [ i ]&&! f [( i +1)/2]) c [ i ]++;
for ( int i =3; i <=100000; i ++) c [ i ]+= c [i -1];
scanf ("%d " ,& N );
while (N--)
{
scanf ("%d %d " ,&L ,& R );
printf ("%d\n" , c [ R ] - c [L -1]);
}
}