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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  map<long long, int> indexes;
  vector<long long> numbers(n);
  for (int i = 0; i < n; ++i) {
    cin >> numbers[i];
    ++indexes[numbers[i]];
  }
  vector<long long> pows_of_two;
  long long base = 1L;
  for (int i = 0; i < 31; ++i) {
    base *= 2;
    pows_of_two.push_back(base);
  }
  long long count = 0;
  for (int i = 0; i < n; ++i) {
    --indexes[numbers[i]];
    for (int j = 0; j < (int)pows_of_two.size(); ++j) {
      long long temp = pows_of_two[j] - numbers[i];
      if ((int)indexes[temp] == 0) continue;
      count += max(0, indexes[temp]);
    }
  }
  cout << count << endl;
  return 0;
}
