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: Wrong Answer)

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<complex>
using namespace std;

vector<int> odd;
vector<int> like2017;
int q;
int ir[100000][2];

void oddmaker(){
odd.push_back(2);
for (int i = 2; i < 100000; i++){
  bool flag = true;
  int sq = sqrt(i);
  sq = max(sq,2);
  for (int k = 2; k <= sq; k++){
    if(i%k == 0){
      flag = false;
      break;
    }
  }
  if(flag) odd.push_back(i);

}
int t = -1;
while(true){
  t++;
  if(odd[t] > 5000) break;
  auto itr = find(odd.begin(), odd.end(), 2*odd[t]-1);
  if( itr != odd.end() ) { like2017.push_back(2*odd[t]-1);
    }
}





}







int main(){

oddmaker();

cin >> q;
for (int i = 0; i < q; i++){
  cin >> ir[i][0] >> ir[i][1];
}

for (int i = 0; i < q; i++){
  int s = 0;
  int e = 0;
  int count = 0;
  for(auto itr = like2017.begin(); itr != like2017.end(); ++itr) {
    count++;
    if (s == 0 && *itr >= ir[i][0]) s = count;
    if (*itr > ir[i][1]){
      e = count;
      break;
    }
  }


cout << e-s << endl;


}

return 0;
}