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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <map>
#include <vector>
#include <deque>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#include <iterator>

using namespace std;

#if 1
#define PRINT(x) { cout << #x << ": " << (x) << endl; }
#define PRINT_CONTAINER(x)  { cout << #x << ": ";                       \
        for (size_t ix = 0; ix < (x).size(); ++ix) cout << (x)[ix] << ", "; \
        cout << endl; }
#else
#define PRINT(x)
#define PRINT_CONTAINER(x)
#endif

#define N 100001
char isprime[N];
void eratosthenes()
{
    for (int i = 2; i < N; i++) isprime[i] = 1;
    for (int i = 2; i < sqrt(N); i++)
    {
        if (isprime[i])
        {
            for (int j = 0; i * (j + 2) < N; j++)
            {
                isprime[i *(j + 2)] = 0;
            }
        }
    }
}

int main(int argc, char *argv[])
{
    eratosthenes();
    int q;
    scanf("%d", &q);
    for (int i = 0; i < q; ++i)
    {
        int l, r;
        scanf("%d %d", &l, &r);
        int cnt = 0;
        for (int j = l; j <= r; j += 2)
        {
            if (isprime[j] && isprime[(j + 1) / 2])
            {
                // printf("%d %d\n", j, (j + 1) / 2);
                ++cnt;
            }
        }
        printf("%d\n", cnt);
    }
}