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;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Program
{
    static void Main(string[] args)
    {
        var flags = sieveAtkin(100000);

        var qNum = int.Parse(Console.ReadLine());
        for(int i = 0; i < qNum; i++)
        {
            var line = Console.ReadLine();
            var min = int.Parse(line.Split(' ')[0]);
            var max = int.Parse(line.Split(' ')[1]);
            var count = 0;
            for (int n = min; n <= max; n += 2)
            {
                if (flags[n] && flags[(n + 1) / 2]) { count++; }
            }
            Console.WriteLine(count.ToString());
        }
        
    }

    public static BitArray sieveAtkin(int n)
    {
        int sqn = (int)Math.Sqrt(n);
        BitArray isp = new BitArray(n+1);
        int b;
        for (int z = 1; z <= 5; z += 4)
        {
            for (int y = z; y <= sqn; y += 6)
            {
                for (int x = 1; x <= sqn && (b = 4 * x * x + y * y) <= n; ++x) isp.Flip(b);
                for (int x = y + 1; x <= sqn && (b = 3 * x * x - y * y) <= n; x += 2) isp.Flip(b);
            }
        }
        for (int z = 2; z <= 4; z += 2)
        {
            for (int y = z; y <= sqn; y += 6)
            {
                for (int x = 1; x <= sqn && (b = 3 * x * x + y * y) <= n; x += 2) isp.Flip(b);
                for (int x = y + 1; x <= sqn && (b = 3 * x * x - y * y) <= n; x += 2) isp.Flip(b);
            }
        }
        for (int y = 3; y <= sqn; y += 6)
        {
            for (int z = 1; z <= 2; ++z)
            {
                for (int x = z; x <= sqn && (b = 4 * x * x + y * y) <= n; x += 3) isp.Flip(b);
            }
        }
        for (int i = isp.NextSetBit(5); i != -1 && i <= sqn; i = isp.NextSetBit(i + 2))
        {
            int zz = i * i;
            for (int k = zz; k <= n; k += zz) isp.Set(k, false);
        }
        isp.Set(2, true); isp.Set(3, true);
        return isp;
    }
}

public static class BitArrayExtension
{
    // 指定ビットを反転する
    public static void Flip(this BitArray array, int index)
    {
        array.Set(index, !array.Get(index));
    }

    public static int NextSetBit(this BitArray array, int index)
    {
        for (int i = index; i < array.Count; i++)
        {
            if (array.Get(i)) { return i; }
        }

        return index;
    } 
}