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;
int main() {
  long long int n;
  cin >> n;
  long long int ar[n];
  for (int i = 0; i < n; i++) cin >> ar[i];
  sort(ar, ar + n);
  map<long long int, long long int> mp;
  long long int prev = ar[0];
  mp[ar[0]] = 1;
  for (int i = 1; i < n; i++) {
    if (ar[i] == prev)
      mp[ar[i]]++;
    else {
      mp[ar[i]] = 1;
      prev = ar[i];
    }
  }
  long long int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    mp[ar[i]]--;
    long long int p = (long long int)(ceil(log2(ar[i])));
    if (ceil(log2(ar[i])) == log2(ar[i])) p++;
    long long int rem = (long long int)(pow(2, p)) - ar[i];
    auto itr = mp.find(rem);
    if (itr != mp.end()) ans += mp[rem];
  }
  cout << ans;
  return 0;
}
