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 n, a[100001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1, greater<int>());
  for (int i = 1; i <= n; i++) {
    int z = ceil(log2(a[i]));
    int y = pow(2, z);
    int x;
    if (a[i] == y)
      x = a[i];
    else
      x = y - a[i];
    for (int j = i + 1; j <= n; j++)
      if (a[j] == x) ans++;
  }
  cout << ans;
  return 0;
}
