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;
const int N = 1e5 + 5;
int a[N];
vector<int> v;
long long sum = 0;
int n;
int main() {
  long long ll = 1;
  for (int i = 1; i < 32; i++) {
    ll *= 2;
    v.push_back(ll);
  }
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < v.size(); j++) {
      if (v[j] <= a[i]) continue;
      sum += upper_bound(a + i + 1, a + n, v[j] - a[i]) -
             lower_bound(a + i + 1, a + n, v[j] - a[i]);
    }
  }
  cout << sum << endl;
  return 0;
}
