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<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include <numeric>
#include<map>
#include<unordered_map>
#include <queue>
 
using namespace std;
using ll=long long;
#define rep(i,n)  for(ll i=0;i<n;++i)
#define all_map(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define ALL(a) (a).begin(),(a).end()

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n)cin >> a[i];
    sort(ALL(a));

    ll ans = 0;
    bool used[n];
    rep(i, n)used[i]=false;
    for(ll i = n-1; i>=0; --i){
        if(used[i])continue;
        for(ll j=40; j>=0; --j){
            ll dif = (1LL<<j)-a[i];
            if(dif==a[i])continue;
            if(dif<0)break;
            ll is = lower_bound(ALL(a), dif)-a.begin();
            if(a[is] == dif&&!used[is]){
                ans++;
                used[i] = true;
                used[is]=true;
                break;
            }
        }
    }
    cout << ans << endl;
}