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 PyPy3 (2.4.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
from pprint import pprint
import math
Q = int(input())
lr = [list(map(int, input().split())) for _ in range(Q)]

# 0以上整数x「未満」の素数をリストに格納して返す
def primes(x):
    if x < 2:
        return []

    primes = [i for i in range(x)]
    primes[1] = 0  # 1は素数ではない

    # エラトステネスのふるい
    for prime in primes:
        if prime > math.sqrt(x):
            break
        if prime == 0:
            continue
        for non_prime in range(2 * prime, x, prime):
            primes[non_prime] = 0

    return [prime for prime in primes if prime != 0]

primes_list = set(primes(10**5+1))
list2017 = [i for i in primes_list if i%2 == 1 and (i+1)//2 in primes_list]
list2017.sort()

def search(n):
    s = 0
    e = len(list2017) - 1
    i = 0
    if n < list2017[s]:
        return 0
    if list2017[e] <= n:
        return len(list2017)
    while s < e:
        i = (s + e)//2
        if list2017[i] == n:
            return i+1
        elif list2017[i] < n:
            s = i
        else:
            e = i

        if s + 1 == e:
            return s + 1


for l,r in lr:
    # pprint([a for a in list2017 if l<=a<=r])
    lidx = search(l-1)
    ridx = search(r)
    # print(ridx)
    # print(lidx)
    print(ridx - lidx)