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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define rng(a) a.begin(),a.end()
using namespace std;
using ll = long long;
using P = pair<int,int>;


int main() {
  int q;
  cin >> q;
  set<int> num;
  for(int i = 1; i <= 100100; i++) {
    bool ok = true;
    for(int j = 2; j*j <=i; j++) {
      if(i%j==0) {
        ok = false;
        break;
      }
    }
    if(ok) num.insert(i);
  }
  rep(xi,q) {
    int l,r, ans = 0;
    cin >> l >> r;
    for(int i = l; i <= r; i++) {
      if(num.count(i)&&num.count((i+1)/2)) ans++;
    }
    cout << ans << endl;
  }
  return 0;
}