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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int asc(const void* a, const void* b)
{
	if (*(int*)a > * (int*)b)return 1;
	if (*(int*)a < *(int*)b)return -1;
	return 0;
}
int main()
{
	int n;
	scanf("%d", &n);
	int i;
	int a[200005];
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	qsort(a, n, sizeof(int), asc);
	int ans = 0;
	int used[200005];
	for (i = 0; i < n; i++)
		used[i] = 0;
	int l, r;
	int p = 1;
	while (p < a[n - 1] + a[n - 2])
		p *= 2;
	while (p > 0)
	{
		l = 0;
		r = n - 1;
		while (l < r)
		{
			if (a[l] + a[r] < p)
				l++;
			else if (a[l] + a[r] > p)
				r--;
			else
			{
				if (used[l] == 0 && used[r] == 0)
				{
					ans++;
					used[l] = 1;
					used[r] = 1;
				}
				l++;
				r--;
			}
		}
		p /= 2;
	}
	printf("%d\n", ans);
	return 0;
}