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 MAX = 1e5 + 10;
int imos[MAX];

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

int main() {
  for (int i = 0; i < MAX; i++) {
    imos[i] = is_prime(i) && is_prime((i + 1) / 2);
  }
  //for (int i = 0; i < 10; i++) cout << i << " = " << imos[i] << endl;
  for (int i = 0; i < MAX - 1; i++) {
    imos[i + 1] += imos[i];
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << imos[r] - imos[l - 1] << endl;
  }
}