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 <cstdio>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int maxn = 1000000 + 20;

bool prime[maxn];

bool isprime(int n)
{
    int t = sqrt(n);
    for(int i = 2; i <= t; i++)
    {
        if(n % i == 0)
        {
            return false;
        }
    }
    return true;
}

int presum[maxn];

void init()
{
    memset(prime, true, sizeof(prime));
    prime[0] = false;
    prime[1] = false;
    for(int i = 2; i < maxn; i++)
    {
        if(!prime[i])
        {
            continue;
        }
        if(isprime(i))
        {
            for(int j = 2; j * i < maxn; j++)
            {
                prime[j * i] = false;
            }
        }
        else
        {
            prime[i] = false;
        }
    }
    for(int i = 1; i < maxn; i++)
    {
        presum[i] = presum[i-1] + (prime[i] && prime[(i+1)/2]);
    }
}

int getsum(int l, int r)
{
    return presum[r] - presum[l - 1];
}

int main()
{
    init();
    int T;
    cin >> T;
    while(T--)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        cout << getsum(l, r) << endl;
    }
    return 0;
}