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)

/*
гонфлад
*/
#pragma ARCENIY_KIRILLOV_UVIDEL_KROKODILOV
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("no-stack-protector")
//#pragma GCC push_options
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC pop_options
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define rng(a) a.begin(), a.end()
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define int long long
using namespace std;
typedef long long ll;

vector<int> a;
vector<bool> used;
int nxt(int a){
    int r = 1;
    int ret = 0;
    while (r <= a){
        r <<= 1;
        ret++;
    }
    return r;
}
int get(int b){
    int n = a.size();
    int l = -1, r = n - 1;
    while (r - l > 1){
        int m = (l + r) / 2;
        if (a[m] < b || (a[m] == b && used[m]))
            l = m;
        else
            r = m;
    }
    return r;
}
main(){
    int n;
    cin >> n;
    a.resize(n);
    used.resize(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(rng(a));
    int ans = 0;
    for (int i = n - 1; i > -1; i--){
            if(used[i]) continue;
        int next_2 = nxt(a[i]);
        int id = get(next_2 - a[i]);
        if (a[id] == next_2 - a[i] && !used[id]){
            used[id] = 1;
            used[i] = 1;
            ans++;
        }
    }
    cout << ans;

}