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: Time Limit Exceeded)

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <vector>

using namespace std;

#define rep(i, n) for (int64_t i = 0; i < (n); i++)
#define irep(i, n) for (int64_t i = 0; i <= (n); i++)
#define rrep(i, n) for (int64_t i = (n)-1; i >= 0; i--)
#define rirep(i, n) for (int64_t i = n; i >= 0; i--)

int main()
{
    int n;
    cin >> n;

    vector<int64_t> a(n);
    multiset<int64_t> s;
    rep(i, n)
    {
        cin >> a[i];
        s.insert(a[i]);
    }

    sort(a.begin(), a.end());

    int result = 0;
    rrep(i, n)
    {
        if (s.count(a[i]) == 0) {
            continue;
        }
        int64_t b = a[i];
        b |= b >> 1;
        b |= b >> 2;
        b |= b >> 4;
        b |= b >> 8;
        b |= b >> 16;
        b |= b >> 32;
        b++;
        int64_t c = b - a[i];
        if ((c != a[i] && s.count(c) > 0) || (s.count(c) > 1)) {
            s.erase(s.find(c));
            s.erase(s.find(a[i]));
            result++;
            //cout << "a=" << a[i] << " c=" << c << endl;
        }
    }
    cout << result << endl;

    return 0;
}