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<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
#define int ll
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;

int gcd(int a, int b){
    if(b==0) return a;
    return gcd(b,a%b);
}
int lcm(int a, int b){
    return a*b/gcd(a,b);
}

vi used;
vector<vi> g;
int dfs(int v){
    int ret = 1;
    used[v]++;
    rep(i,g[v].size()){
        int ne = g[v][i];
        if(used[ne] == 0){
            ret += dfs(ne);
        }
    }
    return ret;
}

signed main(void) {
    int i,j;
    int n;
    cin >> n;
    map<int,vi> ma;
    vi a(n);
    rep(i,n){
        cin >> a[i];
        ma[a[i]].push_back(i);
    }
    g = vector<vi>(n);
    rep(i,n)rep(j,34)if(a[i]<(1LL<<j)){
        int p = (1LL<<j) - a[i];
        rep(k,ma[p].size()){
            int index = ma[p][k];
            if(index != i){
                g[i].push_back(index);
            }
        }
    }
    used = vi(n,0);
    int ans = 0;
    rep(i,n)if(used[i]==0){
        ans += dfs(i)/2;
    }
    cout << ans << endl;
}