output_description: Print the maximum possible number of pairs such that the sum of the integers
  written on each pair of balls is a power of 2.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 ... A_N
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  sys.setrecursionlimit(1000000)
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  AZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # 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)]
  
  
  #####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))))
  
  
  def base_10_to_n_without_0(X, n):
      X -= 1
      if X // n:
          return base_10_to_n_without_0(X // n, n) + [X % n]
      return [X % n]
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  N = I()
  A = IL()
  
  dic = DD(int)
  for a in A:
      dic[a] += 1
  
  kieru = deque([])
  
  A.sort(reverse=True)
  
  ans = 0
  for a in A:
      if kieru and a == kieru[0]:
          kieru.popleft()
          continue
      dic[a] -= 1
      if a == (1 << (a.bit_length() - 1)):
          if dic[a]:
              kieru.append(a)
              dic[a] -= 1
              ans += 1
      else:
          b = (1 << a.bit_length()) - a
          if dic[b]:
              kieru.append(b)
              dic[b] -= 1
              ans += 1
  print(ans)
  
problem_description: Statement
  
  Takahashi has N balls with positive integers written on them. The integer
  written on the i-th ball is A_i. He would like to form some number of pairs
  such that the sum of the integers written on each pair of balls is a power of
  2. Note that a ball cannot belong to multiple pairs. Find the maximum possible
  number of pairs that can be formed.
  
  Here, a positive integer is said to be a power of 2 when it can be written as
  2^t using some non-negative integer t.
samples: [{"input": "3\n    1 2 3", "output": "1\n    \n\nWe can form one pair whose sum of the written numbers is 4 by pairing the\nfirst and third balls. Note that we cannot pair the second ball with itself.\n\n* * *"}, {"input": "5\n    3 11 14 5 13", "output": "2"}]