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)

Q = gets.chomp.to_i
ls, rs = [], []
Q.times do |i|
    ls[i], rs[i] = gets.chomp.split.map(&:to_i)
end
min = ls.min
max = rs.max

is_prime = Array.new(max+1, true)
is_prime[0] = false
is_prime[1] = false

(2..max).each do |i|
    next unless is_prime[i]
    (2..(max/i)).each do |j|
        is_prime[i * j] = false
    end
end

# accum[i]: # of 2017-like num s.t. 1 <= p <= 2i+1
min_i = (ls.min - 1) / 2
max_i = (rs.max - 1) / 2
accum = Array.new(max_i + 1, 0)
(min_i..max_i).each do |i|
    num = 2 * i + 1
    if is_prime[num] && is_prime[(num+1)/2]
        accum[i] = accum[i - 1] + 1
    else
        accum[i] = accum[i - 1]
    end
end

Q.times do |i|
    li = (ls[i] - 1) / 2
    ri = (rs[i] - 1) / 2
    sum = accum[ri]
    sum -= li == 0 ? 0 : accum[li - 1]
    print sum, "\n"
end