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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

const
	num = 100000;
var
	q,l,r,i : Longint;
	not_prime:array[1..num] of Boolean;
	count:array[0..num] of Longint;
function init(): Integer;
var i,j:Longint;
begin
	not_prime[1]:=True;
	for i := 2 to num do
	begin
		if not_prime[i]= False then
		begin
			j:=i+i;
			while j<=num do
			begin
				if not_prime[j]=False then 
					not_prime[j]:=True;
				j:=j+i;
			end;
		end;
	end;
	init := 0;
end;

function get_count(): Integer;
var i:Longint;
begin
	for i := 1 to num do 
	begin
		if  (not_prime[i]=False) and (not_prime[trunc((i+1)/2)]=False) then
		begin
			count[i]:=1;
		end;
		count[i]:=count[i]+count[i-1];
	end;
	get_count := 0;
end;

begin
	init();
	get_count();
	readln(q);
	for i := 1 to q do
	begin
		read(l);
		readln(q);
		writeln(count[q]-count[l-1]);
	end;
	readln;
end.