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: Wrong Answer)

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

namespace AGC029B
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());
            var A = Console.ReadLine().Split().Select(int.Parse).ToList();

            A = A.OrderBy(x => x).ToList();

            var dic = new Dictionary<int, int>();
            for (int i = 0; i < N; i++)
            {
                if (dic.ContainsKey(A[i])) dic[A[i]]++;
                else dic[A[i]] = 1;
            }

            int[] twoPow = new int[30];
            for (int i = 0; i < 30; i++) twoPow[i] = 1 << i;

            int count = 0;
            for (int j = N - 1; j > 0; j--)
            {
                int last = A[j];

                int pair = -1;
                for (int i = 0; i < 30; i++)
                {
                    if (twoPow[i] > last)
                    {
                        pair = twoPow[i] - last;
                        break;
                    }
                }

                if (!dic.ContainsKey(pair)) continue;

                if (dic[pair] > 0 && dic[last] > 0)
                {
                    if (pair != last)
                    {
                        count++;
                        dic[last]--;
                        dic[pair]--;
                    }
                    else
                    {
                        if (dic[pair] >= 2)
                        {
                            count++;
                            dic[last] -= 2;
                        }
                    }
                }
            }

            Console.WriteLine(count);
            Console.ReadKey();
        }
    }
}