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)

/***********************
*  Author: xuziyuan    * 
***********************/
 
#include <bits/stdc++.h>
 
#define rt0 return 0
#define rep(i,n) for(LL i=0;i<n;i++)
#define repn(i,n) for(LL i=1;i<=n;i++)
#define replet(c) for(char c='a';c<='z';c++)
#define LL long long
#define pii pair <LL,LL>
#define pb push_back
#define fi first
#define se second
#define mpr make_pair
#define sqr(a) ((a)*(a))
 
using namespace std;
 
const LL MOD=1e9+7;
 
LL n,a[200010],ans;
bool vis[200010],ok;
vector <LL> pw;
 
LL FIND(LL lb,LL ub)
{
	while(lb<ub)
	{
		LL mid=(lb+ub+1)/2;
		if(vis[mid]) lb=mid+1;
		else ub=mid;
	}
	if(!vis[lb]) return lb;
	return (LL)(-1);
}
 
int main()
{
	pw.pb(1);
	repn(i,40) pw.pb(pw[i-1]*2LL);//,cout<<pw[i]<<endl;
	cin>>n;
	rep(i,n) cin>>a[i];
	sort(a,a+n);
	for(LL i=n-1;i>=0;i--)
	{
		if(vis[i]) continue;
		vis[i]=true;
		LL sum=pw[lower_bound(pw.begin(),pw.end(),a[i])-pw.begin()];
		if(sum==a[i]) sum=pw[lower_bound(pw.begin(),pw.end(),a[i]*2)-pw.begin()];
		LL targ=sum-a[i];
		LL pos=lower_bound(a,a+n,targ)-a,
		pos1=upper_bound(a,a+n,targ)-a;
		if(pos==n||a[pos]!=targ) continue;
		ok=false;
		LL j=FIND(pos,pos1-1);
		if(!ok) vis[i]=false;
		vis[j]=true;
		ans++;
	}
	cout<<ans<<endl;
	rt0;
}