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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Collections;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        AGC029_B solver = new AGC029_B();
        solver.solve(1, in, out);
        out.close();
    }

    static class AGC029_B {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            ArrayList<Long> balls = new ArrayList<Long>();

            for (int i = 0; i < n; i++) {
                balls.add(in.nextLong());
            }

            Collections.sort(balls);

            long tmp;
            long sub;
            long ans = 0;
            ArrayList<Long> already = new ArrayList<Long>();
            for (int i = balls.size() - 1; i > 0; i--) {
                tmp = balls.get(i);
                long binary = (long) Math.floor(Math.log(tmp * 2) / Math.log(2));
                binary = (long) Math.pow(2, binary);
                sub = binary - tmp;
                if (balls.contains(sub) && sub != tmp && !already.contains(sub)) {
                    already.add(sub);
                    ans++;
                    balls.set(i, Long.MAX_VALUE);
                }
            }

            out.print(ans);


        }

    }
}