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: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <utility>
#include <cmath>

using namespace std;

using lli = long long int;
// #define DEBUG
#ifdef DEBUG
#define debug(var) cout << (#var) << ": " << (var) << endl;
template<class T> void dump_array_impl(const T& a) {
  bool first = true;
  for (auto&& x : a) {
    if (!first) { cout << ", "; first = false; }
    cout << x;
  }
}
#define debug_array(a) cout<<(#a)<<": ";dump_array_impl(a);cout<<endl
#else
#define debug(var) {}
#define debug_array(a) {}
#endif

lli n;
vector<lli> v;
int main() {
  cin >> n;
  for (int i=0; i<n; ++i) {
    lli x; cin >> x;
    v.push_back(x);
  }
  sort(begin(v), end(v));
  lli ans = 0;
  for (lli x=1ll<<31; x>=2; x>>=1) {
    vector<lli> vv;
    debug(x);
    for (int i=0; i<v.size(); ++i) {
      lli tgt = x - v[i];
      auto it = lower_bound(begin(v)+i+1, end(v), tgt);
      if (it != end(v) && *it == tgt) {
	v.erase(it);
	++ans;
      } else {
	vv.push_back(v[i]);
      }
    }
    v = move(vv);
  }
  cout << ans << endl;
  return 0;
}