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++ (GCC 9.2.1) (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>;
template<typename T> void chmax(T &a, T b) { a = max(a, b); }
template<typename T> void chmin(T &a, T b) { a = min(a, b); }

// Sieve of Eratosthenes
// https://youtu.be/UTVg7wzMWQc?t=2774
struct Sieve {
    int n;
    vector<int> f, primes;
    Sieve(int n=1):n(n), f(n+1) {
        f[0] = f[1] = -1;
    for (ll i = 2; i <= n; ++i) {
      if (f[i]) continue;
      primes.push_back(i);
      f[i] = i;
      for (ll j = i*i; j <= n; j += i) {
        if (!f[j]) f[j] = i;
      }
    }
  }
  bool isPrime(int x) { return f[x] == x;}
  vector<int> factorList(int x) {
    vector<int> res;
    while (x != 1) {
      res.push_back(f[x]);
      x /= f[x];
    }
    return res;
  }
  vector<P> factor(int x) {
    vector<int> fl = factorList(x);
    if (fl.size() == 0) return {};
    vector<P> res(1, P(fl[0], 0));
    for (int p : fl) {
      if (res.back().first == p) {
        res.back().second++;
      } else {
        res.emplace_back(p, 1);
      }
    }
    return res;
  }
  vector<pair<ll,int>> factor(ll x) {
    vector<pair<ll,int>> res;
    for (int p : primes) {
      int y = 0;
      while (x%p == 0) x /= p, ++y;
      if (y != 0) res.emplace_back(p,y);
    }
    if (x != 1) res.emplace_back(x,1);
    return res;
  }
};

const int MAX_N = 100005;

int main() {
    int q;
    cin >> q;
    Sieve s(MAX_N);
    vector<int> likes;

    for (int p : s.primes) {
        int q = (p+1) / 2;
        if (s.isPrime(q)) likes.push_back(p);
    }

    rep(i, q) {
        int l, r;
        cin >> l >> r;
        int ans = 0;

        for (int like : likes) if (l <= like && like <= r) ans++;
        cout << ans << endl;
    }
    return 0;
}