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 long long int maxn = 1e5;
void runcases(long long int T) {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  map<long long int, long long int> mp;
  vector<long long int> c;
  long long int s = 1;
  while (s < 3e9 * 1LL) {
    c.push_back(s);
    s = s * 2;
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    for (long long int x : c) {
      long long int l = x - v[i];
      if (l < 0) {
        continue;
      }
      ans += mp[l];
    }
    mp[v[i]]++;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T = 1;
  for (long long int t = 1; t <= T; t++) {
    runcases(t);
  }
  return 0;
}
