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() {
  vector<long long int> A;
  long long int n, i, ans, j, P[40], x;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> ans;
    A.push_back(ans);
  }
  sort(A.begin(), A.end());
  for (i = 0, x = 1; i < 32; i++) {
    P[i] = x;
    x = x << 1;
  }
  for (i = 0, ans = 0; i < n; i++) {
    for (j = 0; j < 32; j++) {
      if (binary_search(A.begin() + i + 1, A.end(), P[j] - A[i])) {
        if ((lower_bound(A.begin(), A.end(), P[j] - A[i]) - A.begin()) <= i + 1)
          ans += ((upper_bound(A.begin(), A.end(), P[j] - A[i]) - A.begin()) -
                  i - 1);
        else
          ans += (upper_bound(A.begin(), A.end(), P[j] - A[i]) -
                  lower_bound(A.begin(), A.end(), P[j] - A[i]));
      }
    }
  }
  cout << ans << endl;
}
