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: Runtime Error)

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
{
    static int n;
    static Dictionary<int, int> dict;
    static int [] a;
    static bool[] used;
    
    public static void Main()
    {
	n = int.Parse(ReadLine());
	a = ReadIntArray();
	dict = new Dictionary<int,int>();
	for (int i = 0; i < n; i++) dict.Add(a[i], i);
	used = new bool[n];
	WriteLine(search(0, 0));
    }

    static int search(int i, int ct)
    {
	if (i == n-1){ return ct; }
	if (used[i]){ return search(i+1, ct); }
	
	int x = a[i];

	int m = ct;
	for (int j = 2; j < 31; j++)
	{
	    int p2 = (1 << j);
	    int y = p2 - x;
	    if (dict.ContainsKey(y))
	    {
		int yi = dict[y];
		if (yi != i && !used[yi]){
		    used[i] = used[yi] = true;
		    int nct = search(i+1, ct+1);
		    m = Math.Max(m, nct);
		    used[i] = used[yi] = false;
		}
	    }
	}
	return m;
    }
}