Problem:
Score : 600 points

Problem 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.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

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.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N = gets.to_i
A = gets.split.map(&:to_i).sort

odds = A.select(&:odd?)
evens = A.select(&:even?)

def f(n)
  v = []

  swap = false

  (n.to_s(2).size).times do |i|
    if swap
      if n[i] == 1
        v.push('0')
      else
        v.push('1')
      end
    else
      if n[i] == 1
        swap = true
        v.push('1')
      else
        v.push('0')
      end
    end
  end

  v.reverse.join.to_i(2)
end

def c(list)
  cnt = 0

  counter = Hash.new(0)
  list.each do |l|
    counter[l] += 1
  end

  until list.empty?
    n = list.pop
    next if counter[n] <= 0
    counter[n] -= 1
    t = f(n)
    r = list.bsearch { |x| x >= t }

    if r && r == t && counter[r] > 0
      counter[r] -= 1
      cnt += 1
    end
  end

  cnt
end

puts c(odds) + c(evens)