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;
#define rep(i,n) for (int i=0; i<(int)(n); i++)
using ll = long long;
using vi = vector<int>;
using vl = vector<int>;
using Graph = vector<vector<int>>;  // int <=> ll
using P = pair<int, int>;  // int <=> ll
const int INFTY = 1e9+7;

int main() {
  // 1-indexed
  vector<bool> prime(1e5+1, true);
  vi like(1e5+1, 0);
  prime[1] = false;
  like[1] = 0;

  for (int i=2; i<=1e5; i++) {
    if (prime[i]) {
      for (int j=2*i; j<=1e5; j+=i) {
        prime[j] = false;
      }
    }
  }

  for (int i=3; i<=1e5; i+=2) {
    if (prime[i] && prime[(i+1)/2]) {
      like[i] = like[i-2] + 1;
    } else {
      like[i] = like[i-2];
    }
  }

  int Q;
  cin >> Q;
  rep(i,Q) {
    int l, r;
    cin >> l >> r;
    if (l != 1) l -= 2;
    cout << like[r] - like[l] << endl;
  }
  return 0;
}