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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef double lf;
typedef long double Lf;
typedef pair <int,int> pii;
typedef pair <ll, ll> pll;

#define TRACE(x) cerr << #x << "  " << x << endl
#define FOR(i, a, b) for (int i = (a); i < int(b); i++)
#define REP(i, n) FOR(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define _ << " " <<

#define fi first
#define sec second
#define mp make_pair

const int MAXN = 200100;

int n;

ll p[MAXN];

unordered_map <int, int> cnt;

int main() {
  scanf("%d",&n);
  REP(i, n) {
    scanf("%lld",&p[i]);
    cnt[p[i]]++;
  }
  
  int sol = 0;
  for (int i = 31; i >= 0; i--) {
    ll x = (1LL << i);
    REP(j, n) {
      if (x - p[j] != p[j] && cnt[p[j]] && cnt[x - p[j]]) {
        cnt[p[j]]--;
        cnt[x - p[j]]--;
        sol++;
      }
      while (x - p[j] == p[j] && cnt[p[j]] >= 2) {
        cnt[p[j]] -= 2;
        sol++;
      }
    }
  }

  cout << sol << endl;
  return 0;
}