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>
using namespace std;
#define ll  long long
#define PB push_back
#define ld long double
#define ff first
#define ss second
# define st(v) (v).begin(),(v).end()
#define   pr pair<ll,ll>
const ll N = 1e5+4 , M=N,inf=(ll)1e9;
const ll mod=1e9+7;
ll fastAbs(ll n) { return (n ^ (n >> 31)) - (n >> 31); }
ll multiply(ll a, ll b){ return ((a % mod) * (b % mod)) % mod; }
ll add(ll a, ll b) {     return ((a % mod) + (b % mod)) % mod; }
ll sub(ll a, ll b) { return ((a%mod) - (b % mod)+ mod) % mod ; }
//map< pair<ll,ll> , ll > mp;
const ll dx[4] = {1,0,0,-1};
const ll dy[4] = {0,-1,1,0};
/////////////////////////////////////////////////////
const ll nmax = 2e9 + 5 ;
/////////////////////////////////////////////////////
void solve( ){
     int n ;
     cin >> n ;
     vector < ll > a(n) ;
     for( int i = 0 ; i < n;  i++) cin >> a[i] ;
     sort(st(a));
 	 map < ll , ll > mp ;
 	 for( int i = 0 ; i < n ; i++) mp[a[i]]++;
 	 ll ans =  0 ; 
 	 for( int i = n-1 ; i >=0 ; i--){
 	 	if(mp[a[i]]!=0){
 	 	  ll x = ceil(1.0*log2(a[i]));
 	 	  ll no = a[i];
 	 	  cout << no << " " ;
 	 	  while(no<nmax){
 	 	  		no = powl(2,x) ;
 	 	  		x++;
 	 	  		//cout << no << " " ;
 	 	  		if(mp.find(no-a[i])!=mp.end() && mp[no-a[i]]!=0){
 	 	  			if(no-a[i]==a[i] && mp[a[i]]>=2){
 	 	  				mp[a[i]]-=2;
 	 	  				break ;
 	 	  			}
 	 	  			else if(no-a[i]==a[i] && mp[a[i]]<2 ){
 	 	  				continue ;
 	 	  			}
 	 	  			else{
 	 	  				mp[no-a[i]]--;
 	 	  				ans++;
 	 	  				mp[a[i]]--;
 	 	  				break ;
 	 	  			}
 	 	  		}
 	 	  }
 	 	  //cout << '\n';
  	 	}
 	 }
 	 cout << ans << " " ;
}
int main()
 {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll q=1;
    //cin >> q ;
    while(q--)
      solve();
}