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 <iostream>
#include <string>
using namespace std;

bool is_prime(const unsigned n){
    switch(n){
        case 0: // fall-through
        case 1: return false;
        case 2: // fall-through
        case 3: return true;
    } // n > 3 が保証された

    if(n%2 == 0) return false;
    if(n%3 == 0) return false;
    // 2と3の倍数でないので6の倍数ではないことが保証された

    if(n%6 != 1 && n%6 != 5) return false;
    // 6の倍数前後の数(素数かもしれない数)であることが保証された

    // 6の倍数前後の数を使って試し割りをする
    for(unsigned i=5;i*i<=n;i+=6){
        if(n%i     == 0) return false; // 6n-1
        if(n%(i+2) == 0) return false; // 6n+1
    }

    return true;
}


int main()
{
  int n;
  cin >> n;
  int q[100000];
  for(int i; i<n;i++){
    int l,r;
    cin >> l >> r;
    int count=0;
    for(int i=l; i<=r; i++){
      if(is_prime(i)){
        count++;
      }
    }
    cout << count << "\n";
  }
  return 0;
}