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<bits/stdc++.h>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;A++)

using namespace std;

/////////////////////////////////////////////////////
#define pll pair<ll,ll>


ll N,A[202020];

vector<ll> Av;
int ck[202020];
ll i,j,a,ans;
int main(){
    scanf("%lld",&N);
    for (i = 0; i < N;i++){
        scanf("%lld", &A[i]);
        Av.push_back(A[i]);
    }
    sort(Av.begin(), Av.end());
    for (a = 1; a <= Av[N-1];a*=2)
        ;
    for (i = N - 1; i >= 0; i--)
    {
        if (ck[i])
            continue;
        ck[i] = 1;
        while (a / 2 > Av[i])
        {
            a /= 2;
        }
        ll tg = a-Av[i];
        ll l=0, r=i-1;
        while(l<=r){
            ll c = (l + r) / 2;
            if(Av[c]==tg&&ck[c]==0){
                //printf("%lld ", i);
                ans++;
                ck[c] = 1;
                break;
            }else if(Av[c]<tg){
                l = c + 1;
            }else{
                r = c - 1;
            }
        }
    }

    printf("%lld", ans);
    return 0;
}