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.Arrays;
import java.util.StringTokenizer;
import java.util.Map;
import java.util.HashMap;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * 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;
		FastScanner in = new FastScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		B_PowersOfTwo solver = new B_PowersOfTwo();
		solver.solve(1, in, out);
		out.close();
	}

	static class B_PowersOfTwo {
		public void solve(int testNumber, FastScanner in, PrintWriter out) {
			int n = in.nextInt();
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = in.nextInt();
			}
			Arrays.sort(a);
			Map<Integer, Integer> m = new HashMap<>();
			for (int x : a) {
				addOne(m, x);
			}
			int ans = 0;
			for (int i = n - 1; i >= 0; i--) {
				int x = a[i];
				if (!m.containsKey(x)) {
					continue;
				}
				for (int bit = 31; bit >= 0; bit--) {
					int y = x - (1 << bit);
					if (y < 0) {
						continue;
					}
					if (m.containsKey(y)) {
						removeOne(m, x);
						removeOne(m, y);
						++ans;
					}
					break;
				}
			}
			out.println(ans);
		}

		private void addOne(Map<Integer, Integer> m, int x) {
			int v = 0;
			if (m.containsKey(x)) {
				v = m.get(x);
			}
			m.put(x, v + 1);
		}

		private void removeOne(Map<Integer, Integer> m, int x) {
			int v = m.get(x);
			if (v == 1) {
				m.remove(x);
			} else {
				m.put(x, v - 1);
			}
		}

	}

	static class FastScanner {
		private BufferedReader in;
		private StringTokenizer st;

		public FastScanner(InputStream stream) {
			in = new BufferedReader(new InputStreamReader(stream));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					String rl = in.readLine();
					if (rl == null) {
						return null;
					}
					st = new StringTokenizer(rl);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}
}