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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
namespace kozaiyApp
{
    public class LikeNumber
    {
        static void Main()
        {
            int Q = int.Parse(Console.ReadLine());
            int[] ans = new int[Q];

            for (int i = 0; i < Q; i++)
            {
                string[] input = Console.ReadLine().Split(' ');
                int l = int.Parse(input[0]);
                int r = int.Parse(input[1]);
                int a = 0;

                for (int j = l; j <= r; j++)
                {
                    if (IsLikeNumber(j)) a++;
                }

                ans[i] = a;
            }

            for (int i = 0; i < Q; i++)
            {
                Console.WriteLine(ans[i]);
            }
        }

        // 2017に似た数か
        static bool IsLikeNumber(int n)
        {
            int m = (n + 1) / 2;
            return (n % 2 != 0) && IsPrime(n) && IsPrime(m);
        }

        // 素数判定
        static bool IsPrime(int n)
        {
            for (int i = 2; i * i <= n; i++)
            {
                if (n % i == 0) return false;
            }
            return n != 1;
        }
    }
}