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)

Q = int(input())


# 約数
def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    return divisors
N = 5*10**4
acc = [0]*(N+1)
cnt = 0
for i in range(N+1):
    n1 = 2*i + 1
    n2 = i+1
    div1 = make_divisors(n1)
    div2 = make_divisors(n2)    
    if len(div1) == len(div2) == 2:
        cnt += 1
    acc[i] = cnt
# print(acc[:10])
for i in range(Q):
    l_,r_ = map(int, input().split())
    l = (l_-1)//2
    r = (r_-1)//2
    if l > 0:
        print(acc[r]-acc[l-1])
    else:
        print(acc[r])