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)

# 下記ページを参考に実装している
# https://qiita.com/drken/items/56a6b68edef8fc605821#%E5%95%8F%E9%A1%8C-2-atcoder-abc-084-d---2017-like-number

# エラトステネスのふるいで最大値までの素数をO(1)で判別できるようにする

MAX = 10 ** 5 + 10
is_prime = [True] * MAX
is_prime[0] = False
is_prime[1] = False
for i in range(2, MAX):
  if not is_prime[i]:
    continue
  for j in range(i*2, MAX, i):
    is_prime[j] = False

# 2017と似た数字かどうか
like2017 = [0] * MAX
for i in range(2, MAX):
  if is_prime[i] and is_prime[(i+1)//2]:
    like2017[i] = 1

# 累積和
s = [0] * (MAX+1)
for i in range(MAX):
  s[i+1] = s[i] + like2017[i]
#print(is_prime[:10])
#print(like2017[:10])
#print(s[:10])

# クエリ処理
q = int(input())
for i in range(q):
  l, r = map(int, input().split())
  print(s[r+1] - s[l])