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>

using namespace std;
typedef long long int ll;

#define INF 100000

int main()
{
    int num[100001];
    for(int i = 0; i <= INF; i++) num[i] = 1;
    for(int i = 2; i <= sqrt(INF); i++)
    {
        for(int j = 1; i*(j+1) <= INF; j++)
        {
            if(num[i*(j+1)]) num[i*(j+1)] = 0;
        }
    }
    int ans[INF+1];
    ans[0] = 0;
    ans[1] = 0;
    ans[2] = 0;
    for(int i = 3; i <= INF; i++)
    {
        if(i % 2 && num[i] && num[(i+1)/2]) ans[i] = ans[i-1] +1;
        else ans[i] = ans[i-1];
    }
    int Q; cin >> Q;
    for(int i = 0; i < Q; i++)
    {
        int l,r; cin >> l >> r;
        cout << ans[r] - ans[l-1] << endl;
    }
}