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: Wrong Answer)

include Math

#素数の配列を返すメソッド
def fetch_prime_list(max)
  list = (2..max).to_a
  prime_list = Array.new(max,0)
  sqrt = Math.sqrt(max).floor

  while val = list.shift
    prime_list[val-1] = 1
    if val > sqrt
      break
    end
    list.delete_if{|num| num % val == 0}
  end
    list.each{|num| prime_list[num-1] = 1}
  return prime_list
end

Q = gets.chomp.to_i
query = []
Q.times do
  query.push gets.chomp.split.map(&:to_i)
end

prime_list = fetch_prime_list(100000)
cumsum_list = Array.new(100000,0)
3.step(100000,2) do |n|
  cumsum_list[n-1] = cumsum_list[n-3] + prime_list[n-1]*prime_list[n/2]
end

while q = query.shift
  p cumsum_list[q[1]-1]-cumsum_list[q[0]-3]
end