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 (7.3.0) (judge verdict: Accepted)

import bisect
L = []
for _ in range(int(input())):
  a, b = map(int, input().split())
  a, b = (a-1)//2, (b-1)//2
  L.append([a,b])
X_1 = [i for i in range(2, 10**5+1)]
X_2 = []
while len(X_1)>0:
  X_2.append(X_1[0])
  X_1 = [x for x in X_1 if x%X_1[0] != 0]
T = [0]*(10**5)
for i in range(10**5-1):
  if bisect.bisect_left(X_2, 2*i+1)!=bisect.bisect_right(X_2, 2*i+1)\
  and bisect.bisect_left(X_2, i+1)!=bisect.bisect_right(X_2, i+1):
    T[i+1]=T[i]+1
  else:
    T[i+1]=T[i]
for l in L:
  print(T[l[1]+1]-T[l[0]])