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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int MAX_N = 1e5;
bool is_prime[MAX_N];
int sum[MAX_N];
bool calc_prime(int n) {
  int root = sqrt(n);
  for (int i = 2; i <= root; i++) {
    if (n % i == 0) return false;
  }
  return true;
}

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

int main() {
  is_prime[2] = 1;
  for (int i = 3; i < MAX_N; i += 2) is_prime[i] = calc_prime(i);
  ll cnt = 0;
  for (int i = 1; i <= MAX_N; i += 2) {
    if (like(i)) cnt++;
    sum[i] = cnt;
  }

  int q;
  cin >> q;
  rep(i, q) {
    int l, r;
    cin >> l >> r;
    ll ans = sum[r] - sum[l];
    if(like(l)) ans += 1;
    cout << ans << endl;
  }
  return 0;
}