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)
#define chmin(x,y) x = min(x,y)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int main() {
  int q;
  cin >> q;
  vector<P> A(q), Q(q);
  rep(i,q) {
    int l, r;
    cin >> l >> r;
    A[i] = P(r,l);
    Q[i] = P(l,r);
  }
  sort(A.begin(), A.end(), greater<P>());
  int n = A[0].first;
  vector<bool> isP(n+1, true);
  isP[0] = false;
  isP[1] = false;
  for (int i = 2; i <= sqrt(n); ++i) {
    if (isP[i]) {
      for (int j = 2*i; j <= n; ++j) {
        if (j % i == 0) isP[j] = false;
      }
    }
  }
  vector<int> sum(n+5);
  int temp = 0;
  for (int i = 3; i < n+1; ++i) {
    if (isP[i] && isP[(i+1)/2]) temp++;
    sum[i] = temp;
  }
  for (auto x : Q) {
    int ans = sum[x.second] - sum[x.first-1];
    cout << ans << endl;
  }
  return 0;
}