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 <algorithm>
#include <vector>
#include <unordered_map>
using namespace std;
typedef long long ll;
const long long MOD = 1000000007;

int N;
int AA[200001];
unordered_map<int, int> mp;
bool used[200001];
int ans = 0;

int main() {
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> AA[i];
    }
    sort(AA, AA + N);
    reverse(AA, AA + N);
    for (int i = 0; i < N; i++) {
        if (used[i]) continue;
        int A = AA[i];
        int p = 1 << 30;
        while (p > AA[i]) {
            if (mp.find(p - A) != mp.end()) {
                int id = mp[p - A];
                if (id != i && !used[id] && id >= 0 && AA[id] == p - A) {
                    used[i] = true;
                    used[id] = true;
                    mp[p - A]--;
                    ans += 1;
                    break;
                }
            }
            p /= 2;
        }
    }

    cout << ans << endl;
    return 0;
}