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 list1[100001];
int list2[100001];
int n;
long long f(int m) {
  for (int i = 0; i < n; i++) list2[i] = m - list1[n - 1 - i];
  long long sum = 0;
  long long temp1, temp2;
  for (int i = 0, j = 0; i < n && j < n;) {
    if (list1[i] < list2[j])
      i++;
    else if (list1[i] > list2[j])
      j++;
    else {
      temp1 = 1;
      temp2 = 1;
      while (i + 1 < n && list1[i + 1] == list2[j]) {
        i++;
        temp1++;
      }
      while (j + 1 < n && list2[j + 1] == list1[i]) {
        j++;
        temp2++;
      }
      sum += temp1 * temp2;
      i++;
      j++;
    }
  }
  return sum;
}
int main() {
  cin >> n;
  int max = 0;
  for (int i = 0; i < n; i++) cin >> list1[i];
  sort(list1, list1 + n);
  int mi = 1;
  long long sum = 0;
  for (int i = 0; i < 30; i++) {
    mi *= 2;
    sum += f(mi);
  }
  for (int i = 0; i < n; i++)
    if (list1[i] == (list1[i] & (-list1[i]))) sum--;
  cout << sum / 2;
  return 0;
}
