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)

var primes,sum:array[0..100020] of longint;
var prime:array[0..100020] of boolean;
var n,m,i,j,x,num,y:longint;
begin
  prime[0]:=true;prime[1]:=true;
  for i:=2 to 100000 do
  begin
    if not prime[i] then begin inc(num);primes[num]:=i;end;
    for j:=1 to num do
    begin
      if i*primes[j]>100000 then break;
      prime[i*primes[j]]:=true;
      if i mod primes[j]=0 then break;
    end;
  end;
  fillchar(sum,sizeof(sum),0);
  for i:=1 to 100000 do if (not prime[i]) and (not prime[(i+1) div 2])
  then begin sum[i]:=sum[i-1]+1;//writeln(i);
  end
  else sum[i]:=sum[i-1];
  read(m);
  for i:=1 to m do begin read(x,y);writeln(sum[y]-sum[x-1]);end;
end.