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 int long long
#define PI 3.141592653589793
#define rep(i, n) for (int i = 0; i < (n); i++)
#define REP(i, a, n) for (int i = a; i < (n); i++)
#define rrep(i, n, k) for (int i = (n); i >= (k); i--);
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
const int MOD = 1e9 + 7;
const int INF = 9e18;

signed main(){
    int N;
    cin >> N;
    multiset<int> A;
    for (int i = 0; i < N; i++) {
        int temp;
        cin >> temp;
        A.insert(temp);
    }
    vector<int> twoPow;
    for (int i = 0; i <= 32; i++) {
        twoPow.push_back(1LL<<i);
    }
    reverse(twoPow.begin(), twoPow.end());
    int ans = 0;
    while (!A.empty()) {
        int base;
        {
            auto itr = A.end();
            --itr;
            base = *itr;
        }
        A.erase(base);
        for (auto temp : twoPow) {
            if (temp <= base) break;
            int key = temp - base;
            if (A.find(key) != A.end()) {
                A.erase(key);
                ++ans;
                break;
            }
        }
    }
    cout << ans << endl;
}