You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer x exists so that ai + aj = 2x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

Examples

Input

4
7 3 2 1


Output

2


Input

3
1 1 1


Output

3

Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.
#include <bits/stdc++.h>
const int oo = INT_MAX;
const long long int OO = 1e18;
using namespace std;
long long int GCD(long long int a, long long int b) {
  return ((!b) ? a : GCD(b, a % b));
}
long long int LCM(long long int a, long long int b) {
  return a / (GCD(a, b)) * b;
}
bool isPrime(long long int n) {
  if (n == 2) return 1;
  if (n < 2 || n % 2 == 0) return 0;
  for (long long int i = 3; i * i <= n; i += 2)
    if (n % i == 0) return 0;
  return 1;
}
long long int mod = 1000000000 + 7;
int const N = 1e6;
map<int, int> freq;
long long int numbers[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> numbers[i];
    freq[numbers[i]]++;
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    long long int ele = numbers[i];
    freq[ele]--;
    for (int i = 1; i < 31; ++i) {
      int powOfTwo = 1 << i;
      if (freq[powOfTwo - ele]) ans += freq[powOfTwo - ele];
    }
  }
  cout << ans;
  return 0;
}
