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: Runtime Error)

class Integer
    def divisor_list
        return [1] if self == 1
        (1..Math.sqrt(self).to_i).select do |i|
            self%i == 0
        end.map {|e| [e, self/e]}.flatten.uniq.sort
    end
end
prime=[2,3,5,7]
for i in 8..100010
    next if i%2==0 || i%3==0 || i%5==0 || i%7==0
    prime << i if i.divisor_list.size==2
end
list=[]
for i in 3..100000
    next if i%2==0
    list << i if prime.bsearch{|x|x>=i}==i && prime.bsearch{|x|x>=(i+1)/2}==(i+1)/2
end
Q=gets.to_i
Q.times do
    l,r=gets.split.map(&:to_i)
    a=list.bsearch_index{|x|x>r}
    a=list.size-1 if a.nil?
    puts -list.bsearch_index{|x|x>=l}+a
end