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 Python (3.4.3) (judge verdict: Accepted)

q=int(input())
M=100000

import math
def erato(n):
    prime=[]
    cnt=[0]*n
    limit=math.sqrt(n)
    data=[i+1 for i in range(1,n)]
    while True:
        p=data[0]
        if limit<=p:
            return prime+data
        prime.append(p)
        data=[e for e in data if e%p!=0]

prime=erato(M)
cnt=[0]*M
for i in range(len(prime)):
  cnt[prime[i]]+=0.6
  if 2*prime[i]-1<M:
    cnt[2*prime[i]-1]+=0.6
 
for i in range(1,M):
  if cnt[i]>1:
    cnt[i]=cnt[i-1]//1+1
  else:
    cnt[i]=cnt[i-1]//1
    
#print(cnt)
for i in range(q):
  l,r=map(int,input().split())
  print(cnt[r]-cnt[l-1])