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: Time Limit Exceeded)

import java.util.*;
import java.lang.*;


public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }
        Arrays.sort(a);
        int count = 0;
        for (int i = n - 1; i >= 0; i--) {
            long req = 0;
            if (a[i] == 0) {
                continue;
            }
            for (int j = 0; j < 33; j++) {
                if (Math.pow(2, j) > a[i]) {
                    req = (long)Math.pow(2, j) - a[i];
                    break;
                }
            }
            int min = 0;
            int max = i - 1;
            while (max >= min) {
                if (a[(min + max) / 2] > req) {
                    max = (min + max) / 2 - 1;
                } else if (a[(min + max) / 2] < req) {
                    min = (min + max) / 2 + 1;
                } else {
                    a[i] = 0;
                    a[(min + max) / 2] = 0;
                    count++;
                    Arrays.sort(a);
                    i++;
                    break;
                }
            }
        }
        System.out.println(count);
    }
}