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: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define MM = 1000000000;
#define mod = MM + 7;
#define INF (ll)1e18
#define pi acos(-1.0)
#define MAX 100005
#define NIL -1
#define REP(i, n) for(int (i) = 0; (i < (int)(n)); (i)++)
int main(){
    vector<int> v;
    map<int, int> m;
    int n; cin >> n;
    REP(i, n){
        int a; cin >> a;
        m[a]++;
        v.push_back(a);
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    ll ans = 0; 
    REP(i, n){
        ll a = v[i];
        if(m[a] != 0){
            ll b = 1;
            while(b < a){
                b *= 2;
            }
            if(b == a){
                if(m[a] >= 2){
                    m[a] -= 2;
                    ans++;
                }
            } else {
                ll c = b -a;
                if(m[c] != 0){
                    m[a]--;
                    m[c]--;
                    ans++;
                }
            }
        }
    }
    cout << ans << endl;
}