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<utility>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
#include<string.h>*/
#include<bits/stdc++.h>
#define mp(a,b) make_pair((a),(b))
#define pii pair<int,int>
#define pll pair<LL,LL>
#define pdd pair<double,double>
#define pb push_back
#define x first
#define y second
#define sqr(x) ((x)*(x))
#define EPS 1e-6
#define MEM(x) memset(x,0,sizeof(x))
#define MEMS(x) memset(x,-1,sizeof(x))
#define pi acos(-1)
#define index Index
#define Line pll
using namespace std;
typedef long long LL;
int main(){
    int n;
    scanf("%d",&n);
    multiset<int> s;
    for(int i=0;i<n;i++){
        int x;
        scanf("%d",&x);
        s.insert(x);
    }
    vector<int> v;
    for(int i=0;i<30;i++)
    v.pb(1<<i);
    int ans=0;
    while(!s.empty()){
        int x=*s.rbegin();
        s.erase(s.find(x));
        int target=*upper_bound(v.begin(),v.end(),x);
        int need=target-x;
        if(s.find(need)!=s.end()){
            s.erase(s.find(need));
            ans++;
        }
    }
    printf("%d\n",ans);
}
/*
0 1 0
0 0 1
1 0 1
*/