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 M = 1e9 + 7;
void inp(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void solve() {
  long long n, ans = 0;
  cin >> n;
  long long a[n];
  inp(a, n);
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) m[a[i]]++;
  for (long long i = 0; i < n; i++) {
    for (long long j = 1; j < 40; j++) {
      long long k = pow(2, j) - a[i];
      ans = ans + m[k];
      if (k == a[i]) ans--;
    }
  }
  cout << ans / 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
}
