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)

#pragma region Macros
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
using WGraph = vector<vector<pair<int, ll>>>;

template<class T>inline bool chmax(T &a, const T &b) { if (b > a) { a = b; return true; } return false; }
template<class T>inline bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }

constexpr int dx[4] = {-1, 0, 0, 1}, dy[4] = {0, -1, 1, 0};
constexpr int INF = 1e9;
constexpr long long LINF = 1e18;
constexpr long long MOD = 1e9 + 7;
constexpr double EPS = 1e-10;
constexpr double PI = M_PI;
#pragma endregion

void solve() {
    int n;
    cin >> n;
    vector<ll> a(n);
    for (int i=0; i<n; i++) {
        cin >> a.at(i);
    }

    sort(a.begin(), a.end());
    map<int, int> mp;
    for (int i=0; i<n; i++) {
        mp[a.at(i)]++;
    }
    int res = 0;
    for (int i=n-1; i>=0; i--) {
        if (mp[a.at(i)] == 0) continue;
        int tmp = 1;
        while (tmp < a.at(i)) {
            tmp *= 2;
        }
        int dif = tmp - a.at(i);
        if (mp[dif] > 0) {
            mp[dif]--;
            res++;
        }
    }

    cout << res << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    solve();

    return 0;
}