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;
map<int, int> occ;
vector<int> even, odd, neven, nodd;
long long ans;
int n;
bool isp2(int x) { return (!(x & (x - 1)) && x); }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  ans = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (occ.count(1 - x)) ans += occ[1 - x];
    if ((abs(x)) & 1) {
      int sz = odd.size();
      for (int j = 0; j < sz; j++) {
        int y = odd[j];
        if (isp2(x + y)) ans += occ[y];
      }
      if (!occ.count(x)) odd.push_back(x);
      occ[x]++;
    } else {
      int sz = even.size();
      for (int j = 0; j < sz; j++) {
        int y = even[j];
        if (isp2(x + y)) {
          ans += occ[y];
        }
      }
      if (!occ.count(x)) even.push_back(x);
      occ[x]++;
    }
  }
  cout << ans;
  return 0;
}
