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 Ruby (2.3.3) (judge verdict: Accepted)

require "prime"
q=gets.chomp.to_i
a=Array.new(q)

q.times do |i|
  l,r=gets.chomp.split(" ").map { |e| e.to_i }
  a[i] = {"l" => l,"r" => r }
end

b = Array.new(10**5 + 1,0)

def judge_prime_numeber(n)

  a = n.prime_division
  if a.length != 1
    return false
  else
    if a[0][0] != n
        return false
    else
      return true
    end
  end
end


for i in 1..10**5 do
  if judge_prime_numeber(i) && judge_prime_numeber((i+1)/2)
    b[i] += b[i-1] + 1
  else
    b[i] += b[i-1]
  end
end

for i in 0..q-1 do
  l=a[i]["l"]
  r=a[i]["r"]
  puts b[r] - b[l-1]
end