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 (ll i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;
template <typename T> using vec = std::vector<T>;

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<pair<int, int>> factor(int x) {
        vector<int> fl = factorList(x);
        if (fl.size() == 0) return {};
        vector<pair<int, int>> res(1, pair<int, int>(fl[0], 0));
        for (int p : fl) {
            if (res.back().first == p) {
                res.back().second++;
            } else {
                res.emplace_back(p, 1);
            }
        }
        return res;
    }
};


int main() {
    int Q;
    cin >> Q;
    Sieve sieve(1e5+10);
    
    vec<int> L2017(1e5+100,0);
    for(int i=1;i<=1e5;i+=2){
        if(i%2==0)continue;
        if(sieve.isPrime(i) && sieve.isPrime((i+1)/2)) L2017[i] = 1;
    }

    rep(i,1e5) L2017[i+1]+=L2017[i];

    rep(i,Q){
        int l, r;
        cin >> l >> r;
        cout << L2017[r]-L2017[l-1] << '\n';
    }

}