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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace AtCoder
{
    class Program
    {
        private static string Read() { return Console.ReadLine(); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }

        static int Partner(int n)
        {
            long N = 2;
            while (N <= n) { N <<= 1; }
            return (int)(N - n);
        }

        static int Solve(int N, int[] A)
        {
            var counts = new Dictionary<int, int>();
            foreach (int a in A)
            {
                if (!counts.ContainsKey(a)) { counts[a] = 0; }
                ++counts[a];
            }
            int ans = 0;
            foreach (int i in counts.Keys.OrderByDescending(i => i).ToArray())
            {
                int j = Partner(i);
                while (counts[i] > 0)
                {
                    --counts[i];
                    if (counts.ContainsKey(j) && counts[j] > 0)
                    {
                        --counts[j];
                        ++ans;
                    }
                }
            }
            return ans;
        }

        static void Main(string[] args)
        {
            int N = ReadInt();
            int[] A = ReadInts();
            Console.WriteLine(Solve(N, A));
        }
    }
}