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;

const int MAXN = 1e5;
vector<bool> isprime(MAXN + 10, true);

bool f(int n) {
  return isprime[n] && isprime[(n + 1) / 2];
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int Q;
  cin >> Q;
  isprime[0] = false, isprime[1] = false;
  for (int i = 2; i * i <= MAXN; i++) {
    if (!isprime[i]) continue;
    int k = 2;
    while (i * k <= MAXN) {
      isprime[i * k++] = false;
    }
  }
  vector<int> cs(MAXN + 10, 0);
  for (int i = 1; i <= MAXN; i++) {
    if (f(i)) cs[i]++;
    cs[i + 1] += cs[i];
  }
  for (int i = 0; i < Q; i++) {
    int l, r;
    cin >> l >> r;
    cout << cs[r] - cs[l - 1] << '\n';
  }
  return 0;
}