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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.Comparator;
import java.util.TreeMap;

public class Main {

    private int n;
    private long[] a;
    
    public static void main(String[] args) {
	Main m = new Main();
	m.solve();
    }

    private void set() {
	try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
	    n = Integer.parseInt(br.readLine());
	    a = Arrays.stream(br.readLine().split(" ")).mapToLong(Long::parseLong)
		.map(i -> -i).sorted().map(i -> -i).toArray();
	} catch(IOException e) {
	    e.printStackTrace();
	}
    }

    private void solve() {
	set();
	int pair = 0;
	double loge2 = Math.log(2);
	for (int i=0;i<n;i++) {
	    if (a[i] < 0) continue;
	    int t = (int)(Math.log(a[i])/loge2) + 1;
	    long p = (1 << t) - a[i];
	    for (int j=i+1;j<n;j++) {
		if (a[j] < 0) continue;
		if (a[j] == p) {
		    a[j] = -1;
		    pair++;
		    break;
		}
	    }
	}
	System.out.println(pair);
    }
}