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;

bool is_prime(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}

bool like_2017(int n) {
  if (n % 2 == 0) return false;
  return is_prime(n) and is_prime((n + 1) / 2);
}

int main() {
  int Q;
  cin >> Q;
  vector<int> L(Q), R(Q);
  for (int i = 0; i < Q; i++) {
    cin >> L[i] >> R[i];
  }
  vector<int> P(100005);
  for (int i = 2; i <= 100000; i++) {
    P[i] = like_2017(i);
    P[i] += P[i - 1];
  }
  for (int i = 0; i < Q; i++) {
    cout << P[R[i]] - P[L[i] - 1] << endl;
  }

  return 0;
}
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