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.

Examples

Input

3
1 2 3


Output

1


Input

5
3 11 14 5 13


Output

2
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.AbstractMap;
import java.util.TreeMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Comparator;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
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);
        BPowersOfTwo solver = new BPowersOfTwo();
        solver.solve(1, in, out);
        out.close();
    }

    static class BPowersOfTwo {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // input
            int n = in.nextInt();
            TreeMap<Integer, Integer> mp = new TreeMap<>(Comparator.naturalOrder());
            for (int i = 0; i < n; i++) {
                mp.merge(in.nextInt(), 1, Integer::sum);
            }
            int ans = 0;
            while (!mp.isEmpty()) {
                int use = mp.lastKey();
                mp.merge(use, -1, Integer::sum);
                if (mp.get(use) == 0) {
                    mp.remove(use);
                }
                int target = 1;
                while (target <= use) {
                    target *= 2;
                }
                target -= use;
                if (mp.containsKey(target)) {
                    mp.merge(target, -1, Integer::sum);
                    if (mp.get(target) == 0) {
                        mp.remove(target);
                    }
                    ans++;
                }


            }
            out.println(ans);
        }

    }
}

