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 int OO = 0xfffffff;
const int _OO = -1 * OO;
const double EPS = (1e-9);
const double pi = 3.14159265;
using namespace std;
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, -1, 1};
int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
int dy8[8] = {1, -1, 0, 1, -1, 1, -1, 0};
const int N = 1e5 + 7;
map<string, bool> mp;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  vector<long long> v(30);
  for (int i = 1; i <= 30; i++) v[i - 1] = pow(2, i);
  map<long long, long long> cnt;
  long long n, ans = 0;
  cin >> n;
  long long a[n];
  for (int i = 0; i < (int)(n); ++i) cin >> a[i], cnt[a[i]]++;
  for (int i = 0; i < n; i++) {
    long long cur = a[i];
    for (int ii = 0; ii < 30; ii++) {
      if (cnt[v[ii] - cur] != 0) {
        ans += cnt[v[ii] - cur];
        if (v[ii] - cur == cur) ans--;
      }
    }
  }
  cout << (ans >> 1);
  return 0;
}
