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.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		
		Long[] A = new Long[N];
		HashMap<Long, Integer> map = new HashMap<Long, Integer>();
		for(int i=0;i<N;i++) {
			A[i] = sc.nextLong();
			if (map.containsKey(A[i])) map.put(A[i],1+map.get(A[i]));
			else map.put(A[i],1);
 		}
		Arrays.sort(A,Collections.reverseOrder());
		int ret = 0;
		for(int i=0;i<N;i++) {
			if(map.get(A[i])==0) continue;
			map.put(A[i],map.get(A[i]-1));
			//一番左のbitを１にしたのを２倍
			long k2 = 2*Long.highestOneBit(A[i]);
			long diff = k2-A[i];
			if(map.containsKey(diff) && map.get(diff)>0) {
				map.put(diff,map.get(diff)-1);
				ret++;
			}
			System.out.println(ret);
		}
	}
}