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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define SC scanf
#define PR printf
#define pb push_back
#define mp make_pair
#define SZ(v) ((int)(v).size())
#define rep(i,n) for(int i=0;i<(n);i++)
#define INF 1000000000
#define ll long long
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef pair<int,pii> pip;
typedef pair<pii,pii> ppp;
typedef vector<int> vi;
typedef vector<pii> vii;
const int xx[4]={1,-1,0,0};
const int yy[4]={0,0,1,-1};
int n;
map<ll,int>cnt;
ll pow2[50];
int main(){
    ios::sync_with_stdio(false);
    pow2[0]=1;
    for(int i=1;i<=30;i++){
    	pow2[i]=pow2[i-1]*2;
//    	cout<<pow2[i]<<' ';
	}
//	cout<<endl;
	cin>>n;
	vector<ll>a(n+1,0);
	for(int i=0;i<n;i++){
		cin>>a[i];
		cnt[a[i]]++;
	}
	sort(a.begin(),a.end());
	reverse(a.begin(),a.end());
	int ans=0;
	for(int i=0;i<n;i++){
		int pos=0;
		while(a[i]>pow2[pos]){
			pos++;
		}
		if(pow2[pos]==a[i]){
			pos++;
			if(cnt[a[i]]>1){
				ans++;
				cnt[a[i]]-=2;
			}
		}
		else{
			ans++;
			cnt[a[i]]--;
			cnt[pow2[pos]-a[i]]--;
		}
	}
	cout<<ans<<endl;
	return 0;
}