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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        ArrayList<Long> a = new ArrayList<Long>();
        for(int i=0; i<n; i++){
            long tmp = sc.nextLong();
            a.add(tmp);
        }
        sc.close();

        int npow = 32;
        long[] pow2 = new long[npow];
        for(int i=0; i<pow2.length; i++){
            pow2[i] = (long)Math.pow(2, i);
        }

        Collections.sort(a);

        int ans = 0;
        out: for(int i=a.size()-1; i>=0; i--){
            long cur = a.get(i);
            if(i > 0){
                long cur2 = a.get(i-1);
                if(cur == cur2){
                    for(int j=0; j<npow; j++){
                        if(cur*2 == pow2[j]){
                            ans++;
                            i--;
                            continue out;
                        }
                    }
                }
            }
            for(int j=0; j<npow; j++){
                if(cur >= pow2[j]) continue;
                int pos = Collections.binarySearch(a, pow2[j]-cur, (ma, mb) -> Long.compare(ma, mb));
                if(pos >= 0 && pos < i){
                    a.remove(i);
                    a.remove(pos);
                    i--;
                    ans++;
                    break;
                }
                break;
            }
        }

        System.out.println(ans);
    }
}