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, i, j;
  vector<long long int> vec;
  unordered_map<long long int, long long int> mp;
  j = 2;
  for (i = 0; i < 45; i++) {
    vec.push_back(j);
    j = j * 2;
    if (j >= 1e10) break;
  }
  cin >> n;
  long long int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (mp.find(a[i]) == mp.end())
      mp[a[i]] = 1;
    else
      mp[a[i]]++;
  }
  long long int count = 0;
  for (i = 0; i < n; i++) {
    mp[a[i]]--;
    for (j = 0; j < vec.size(); j++) {
      if (vec[j] - a[i] > 0 && mp.find(vec[j] - a[i]) != mp.end())
        count += mp[vec[j] - a[i]];
    }
  }
  cout << count << endl;
  return 0;
}
