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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int main()
{
  int Q;scanf("%d",&Q);
  bool is_prime[100001],alike[100001];
  fill(is_prime,is_prime+100001,true);fill(alike,alike+100001,false);
  is_prime[0]=false,is_prime[1]=false;
  for(int i=2;i<100001;i++)
  {
    if(!is_prime[i])continue;
    for(int j=i*2;j<100001;j+=i)is_prime[j]=false;
  }
  for(int i=3;i<100001;i+=2)if(is_prime[(i+1)/2]&&is_prime[i])alike[i]=true;
  int count_prime[100001];fill(count_prime,count_prime+100001,0);
  for(int i=3;i<100001;i+=2)count_prime[i]=count_prime[i-2]+alike[i];
  for(int l,r,i=0;i<Q;i++)
  {
    scanf("%d %d",&l,&r);
    int ans=count_prime[r]-count_prime[l];
    if(alike[l])ans++;
    printf("%d\n",ans);
  }
}