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;
void inout() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  inout();
  long long int n, i;
  cin >> n;
  long long int a[n];
  map<long long int, long long int> mp;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  long long int ans = 0;
  for (auto j : mp) {
    for (long long int x = 0; x <= 31; x++) {
      if (pow(2LL, x) - j.first > j.first) {
        long long int z = pow(2LL, x) - j.first;
        if (mp.find(z) != mp.end()) {
          ans += (mp[z] * mp[j.first]);
        }
      } else if (pow(2LL, x) - j.first == j.first) {
        long long int z = pow(2LL, x) - j.first;
        if (mp.find(z) != mp.end()) {
          ans += ((mp[z] - 1) * (mp[z]) / 2);
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}
