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: Runtime Error)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        int[][] masu = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i >= j) {
                    masu[i][j] = 0;
                } else {
                    masu[i][j] = a[i] + a[j];
                    // System.out.println(masu[i][j]);
                }
            }
        }
        long ans = 0;
        long count = 0;
        long max = a[n - 1] + a[n - 2];
        long nokori = 0;
        boolean flag = true;
        long kazu = n * (n - 1) / 2;
        long cnt = 0;
        while (flag) {
            count = 0;
            cnt = 0;
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (masu[i][j] == 2) {
                        count++;
                    }
                    if (masu[i][j] % 2 == 1) {
                        masu[i][j] = 0;
                    }
                    masu[i][j] = masu[i][j] / 2;
                    // System.out.println(masu[i][j]);
                    if (masu[i][j] == 0) {
                        cnt++;
                    }
                }
            }
            if (ans < count) {
                ans = count;
            }
            if (kazu == cnt) {
                flag = false;
            }
        }
        System.out.println(ans);
    }
}