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 Python (3.4.3) (judge verdict: Accepted)

#https://qiita.com/masashi127/items/88436e16e3672b387a58
#--------------------------------------------
import math

def sieve_of_erastosthenes(num):
    input_list = [False if i % 2 == 0 or i % 3 == 0 or i % 5 == 0  else True for i in range(num)]
    input_list[0] = input_list[1] = False
    input_list[2] = input_list[3] = input_list[5] = True
    sqrt = math.sqrt(num)

    for serial in range(3, num, 2):

        if serial >= sqrt:
            return input_list

        for s in range(serial ** 2, num, serial): 
            input_list[s] = False

n=100000
input_list = sieve_of_erastosthenes(n)
prime_list = [i for i, b in enumerate(input_list) if b == True]
#print(prime_list)
#print(sum(prime_list))

#--------------------------------------------

count=[0 for i in range(100002)]
for i in prime_list:
  count[i]=1
#print(count)

nitakazu_count=[0 for i in range(50001)]
for i in range(50001):
  if count[2*i+1]==1 and count[i+1]==1:
    nitakazu_count[i]=1
  
  
ruiseki_count=[0 for i in range(50001)]
for i in range(50000):
  ruiseki_count[i+1]=ruiseki_count[i]+nitakazu_count[i+1]
#print(ruiseki_count)

nn=int(input())

for i in range(nn):
  aa,bb=[int(i) for i in input().split()]
  print(ruiseki_count[int((bb-1)/2)]-ruiseki_count[max(int((aa-1)/2)-1,0)])