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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

#define ll  long long
#define PR  pair<int, int>

const int N = 3e5 + 10;
const int MOD = 1e9 + 7;
const long long INF = 5e18;
const double PI = 2 * acos(0.0);
//template<typename T> using ordered_set = tree<T, nulong long_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    vector<int> p2(31, 1);
    for (int i = 2; i < 31; i++) {
        p2[i] = p2[i - 1] * 2;
    }
    int n;
    cin >> n;
    multiset<int> ms;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        ms.insert(x);
    }
    int ans = 0;
    while (!ms.empty()) {
        int a = *ms.rbegin();
        ms.erase(ms.lower_bound(a));
        for (int i = 30; i >= 0; i--) {
            int need = p2[i] - a;
            if (need < 0) {
                break;
            }
            if (ms.find(need) != ms.end()) {
                ms.erase(ms.lower_bound(need));
                ans++;
                break;
            }
        }
    }
    cout << ans << '\n';
    return 0;
}