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;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long cnt = 0;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long st = 1;
    while (st < a[i]) st *= 2;
    long long cntt = 0;
    while (st <= 20000000000) {
      cntt += mp[st - a[i]];
      st *= 2;
    }
    ans += cntt;
    mp[a[i]]++;
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
