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 a[100100];
bool check(long long a) {
  bool ans = false;
  if (a != 0) {
    if ((a & (a - 1)) == 0) ans = true;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long i, j, n;
  bool var;
  map<long, long> se;
  map<long, long>::iterator it;
  long long fo, sum, ct = 0, tem;
  scanf("%ld", &n);
  for (i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
  }
  se.insert(make_pair(a[0], 1));
  for (i = 1; i < n; i++) {
    for (int po = 1; po <= 31; po++) {
      tem = 1 << po;
      sum = tem - a[i];
      if (sum < 0) {
        continue;
      } else {
        fo = se.count(sum);
        if (fo == 1) {
          it = se.find(sum);
          ct += it->second;
        }
      }
    }
    if (se.count(a[i]) == 0) {
      se.insert(make_pair(a[i], 1));
    } else {
      se[a[i]]++;
    }
  }
  cout << ct << endl;
  return 0;
}
