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<algorithm>
#include<cmath>
#include<iomanip>
#include<iostream>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
#include<vector>
using uint = unsigned int;
using ll = long long;
enum : int { M = (int)1e9 + 7 };
enum : ll { MLL = (ll)1e18L + 9 };
using namespace std;
#ifdef LOCAL
#include"rprint2.hpp"
#include"debug_deque.hpp"
#define vector DebugDeque
#else
#define FUNC(name) template <ostream& out = cout, class... T> void name(T&&...){ }
FUNC(prints) FUNC(printe) FUNC(printw) FUNC(printew) FUNC(printb) FUNC(printd) FUNC(printde);
#endif

#include<unistd.h>
struct Buffer{
    int idx;
    enum { chunk = 1 << 16 };
    char buf[chunk];
    Buffer(){ init(); }
    void init(){ idx = 0; read(0, buf, chunk); }
    void check(){ if(idx == chunk){ init(); } }
    int next(){
        static char notDigits[256] = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1, 0,0,0,0,0,0,0,0,0,0,};
        int ret = 0, sign = 1;
        while(check(), notDigits[(int)buf[idx]]){ idx++; }
        if(check(), buf[idx] == '-'){ sign = -1; idx++; }
        while(check(), !notDigits[(int)buf[idx]]){ ret = ret * 10 + buf[idx++] - '0'; }
        return sign * ret;
    }
};

string ans;
void out(int num){
    if(!num){
        // putchar('0');
        ans += '0';
        return;
    }
    char buf[16];
    int idx = 0;
    while(num){
        auto ret = div(num, 10);
        buf[idx] = ret.rem + '0';
        num = ret.quot;
        idx++;
    }
    for(int i = idx - 1; i >= 0; i--){
        // putchar(buf[i]);
        ans += buf[i];
    }
}

int main(){
    // cin.tie(0);
    // ios::sync_with_stdio(false);
    Buffer buf;
    const int lim = 1e5;
    vector<char> prime(lim + 1, true);
    prime[0] = prime[1] = false;
    for(int i = 2; i <= lim; i++){
        if(prime[i]){
            for(int j = i + i; j <= lim; j += i){
                prime[j] = false;
            }
        }
    }
    vector<int> accs(lim + 1);
    for(int i = 1; i <= lim; i++){
        accs[i] = accs[i - 1] + (i & 1 && prime[i] && prime[(i + 1) / 2]);
    }
    // printd(accs[lim]);
    // int q; cin >> q;
    int q = buf.next();
    // stringstream ss;
    ans.reserve(4 * q);
    for(int i = 0; i < q; i++){
        // int l, r;
        // cin >> l >> r;
        int l = buf.next(), r = buf.next();
        // cout << accs[r] - accs[l - 1] << '\n';
        // ss << accs[r] - accs[l - 1] << '\n';
        out(accs[r] - accs[l - 1]);
        // putchar('\n');
        ans += '\n';
    }
    // cout << ss.str();
    cout << ans;
    // write(1, ans.data(), ans.size());
}