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)

import math
prime = [i for i in range(2, 10**5+1)]
for i in range(2, math.floor(math.sqrt(10**5))+1):
    prime = [p for p in prime if (p == i or p % i != 0)]
isPrime = [False for i in range(10**5+1)]
for p in prime:
    isPrime[p] = True
cntPrime = [0 for i in range(10**5+1)]
for i in range(2, 10**5+1):
    if i % 2 == 1 and isPrime[i] and isPrime[(i+1)//2]:
        cntPrime[i] = cntPrime[i-1]+1
    else:
        cntPrime[i] = cntPrime[i-1]

Q = int(input())
for i in range(Q):
    l, r = [int(_) for _ in input().split()]
    print(cntPrime[r]-cntPrime[l-1])