output_description: Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th
  query.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      Q
      l_1 r_1
      :
      l_Q r_Q
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  from collections import defaultdict
  from collections import deque
  from functools import lru_cache
  
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  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
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primes = []
      while seachList[0] <= max:
          primes.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primes.extend(seachList)
      return primes
  
  
  P = make_primes(100000)[1:]
  like = []
  for p in P:
      if len(factorization((p + 1) // 2)) == 1 and factorization((p + 1) // 2)[0][1] == 1:
          like.append(p)
  
  imosu = [0 for i in range(101000)]
  for i in like:
      imosu[i] += 1
      imosu[i + 1] -= 1
  
  ans = [0]
  for i in imosu:
      ans.append(ans[-1] + i)
  
  ruiseki = [0]
  for i in ans:
      ruiseki.append(ruiseki[-1] + i)
  Q = I()
  for _ in range(Q):
      l, r = IL()
      print(ruiseki[r + 2] - ruiseki[l + 1])
  
problem_description: 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.
samples: [{"input": "1\n    3 7", "output": "2\n    \n\n  * 3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.\n  * 5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.\n  * 7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.\n\nThus, the response to the first query should be 2.\n\n* * *"}, {"input": "4\n    13 13\n    7 11\n    7 11\n    2017 2017", "output": "1\n    0\n    0\n    1\n    \n\nNote that 2017 is also similar to 2017.\n\n* * *"}, {"input": "6\n    1 53\n    13 91\n    37 55\n    19 51\n    73 91\n    13 49", "output": "4\n    4\n    1\n    1\n    1\n    2"}]