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 static System.Console;
using System.Runtime.CompilerServices;
using static MyUtil;

class MyUtil
{
    public static int[] ReadIntArray()
    {
        return ReadLine().Split().Select(x => int.Parse(x)).ToArray();
    }
}

class Program
{
    public static void Main()
    {
        int n = int.Parse(ReadLine());
        int[] a = ReadIntArray();
        Array.Sort(a);
        bool[] used = new bool[n];

        int ans = 0;
        for (int i = n-1; i > 0; i--)
        {
            if (used[i]) continue;
            int x = a[i];
            for (int j = 2; j < 31; j++)
            {
                int p2 = (1 << j);
                int y = p2 - x;
                int idx = Array.BinarySearch(a, y);

                if (idx >= 0 && idx < i)
                {
                    while (used[idx] && idx < n-1 && a[idx]==a[idx+1]) idx++;
                    if (!used[idx])
                    {
                        ans++;
                        used[i] = used[idx] = true;
                        break;
                    }
                }
            }
        }

        WriteLine(ans);
    }
}