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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

#define IS_SQUARE(d) (((d) & ((d) - 1)) == 0)

int compare_int(const void *a, const void *b)
{
	// 降順 
	return *(int *) b - *(int *) a;
}

int main(int argc, char *argv[])
{
	int n, i, c;
	int j;
	int a[200000];
	
	int x, y;
	int l_x, l_y;
	
	scanf("%d", &n);

	for (i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	
	qsort(a, n, sizeof(int), compare_int);

	c = 0;
	l_x = 0;
	l_y = 0;

	for (i = 0; i < n; i++) {
		x = a[i];
		if (x == 0) {continue;}
		j = i + 1;
		if ((l_x == x)&&(l_y > i)) {
			j = l_y + 1;
		}
		for (; j < n; j++) {
			y = a[j];
			if (y == 0) {continue;}
			if (IS_SQUARE(x + y)) {
				a[j] = 0;
				c++;
				break;
			}
		}
		l_x = x;
		l_y = j;
	}

	printf("%d\n", c);
	
	return 0;
}