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)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;

//
#define int long long
#define pb(x) push_back(x)
#define m0(x) memset((x), 0, sizeof(x))
#define mm(x) memset((x), -1, sizeof(x))

//container
#define ALL(x) (x).begin(), (x).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define UNIQUE(v) (v).erase(unique((v).begin(), (v).end()), (v).end());
#define PERM(c) \
  sort(ALL(c)); \
  for (bool c##p = 1; c##p; c##p = next_permutation(ALL(c)))

// debug
#define GET_VAR_NAME(variable) #variable
#define test(x) cout << GET_VAR_NAME(x) << " = " << x << endl;

// bit_macro
#define bit(n) (1LL << (n))
#define bitset(a, b) (a) |= (1 << (b))
#define bitunset(a, b) (a) |= ~(1 << (b))
#define bitcheck(a, b) ((((a) >> (b)) & 1) == 1)
#define bitcount(a) __builtin_popcountll((a))

//typedef
typedef long long lint;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

//constant
const int INF = (int)1e18;
const int MOD = (int)1e9 + 7;
const double EPS = (double)1e-10;
const int dx[] = {-1, 0, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, 0, 1, -1, 1, -1};

//
template <typename T>
void chmax(T &a, T b) { a = max(a, b); }
template <typename T>
void chmin(T &a, T b) { a = min(a, b); }
//
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
struct Accelerate_Cin {
  Accelerate_Cin() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
  };
};

int N, A[202020];
int mp[21][2020200];

signed main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }

  sort(A, A + N, greater<int>());

  for (int i = 1; i <= 20; i++) {
    for (int j = 0; j < N; j++) {
      int tmp = pow(2, i);
      if (A[j] < tmp) break;
      if (A[j] >= tmp * 2) break;

      //A[i]=2^i+k(0<=k<2^i)で表されるときの、kの個数
      mp[i][A[j] % tmp]++;
    }
  }

  int ans = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 2; j <= 20; j++) {
      int tmp = pow(2, j - 1);
      if (tmp == A[i] && mp[j - 1][0] == 1) continue;
      if (tmp - A[i] < 0) continue;
      ans += mp[j - 1][tmp - A[i]];
    }
  }

  cout << ans << endl;
  return 0;
}