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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 
int sort(const void *A,const void *B){return *(long long*)A>*(long long*)B?1:0;}
 
int main(void){
    int n;
    scanf("%d",&n);
    long long a[n],v[n],ans=0;
    for (int i=0; i<n; i++){
        scanf("%lld",&a[i]);
        v[i]=0;
    }
    qsort (a,n,sizeof(long long),sort);
    for (int i=31; i>=0; i--){
        long long count=0,back=n-1,sum=pow(2,i);
        while(count<back){
            if (v[count]==1){
                count++;
            } else if (v[back]==1){
                back--;
            } else if (a[count]+a[back]==sum){
                v[count]=1;
                v[back]=1;
                ans++;
                count++;
                back--;
            } else if (a[count]+a[back]<sum){
                count++;
            } else {
                back--;
            }
        }
    }
    printf("%lld\n",ans);
	return 0 ;
}