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: Time Limit Exceeded)

q = gets.to_i
lr = q.times.map{gets.chomp.split(" ").map(&:to_i)}
m = lr.flatten.max
n = Array.new(m+1,1)
n[0..1] = [0,0]
p = 0
Math.sqrt(m).floor.times do
  p = n.index(1)
  n[p] = 2
  for i in p..m/p
    n[i*p] = 0
  end
end
for i in 0..m
  if n[m-i] != 0 && n[(m-i+1)/2] != 0
    n[m-i] = 0
    n[m-i..m] = n[m-i..m].map{|num|num += 1}
  else
    n[m-i] = 0
  end
end
for i in 0..q-1
  puts n[lr[i][1]] - n[lr[i][0]-1]
end