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;
vector<long long int> v;
vector<long long int> v1;
vector<long long int> v2;
const long long int inf = 100000000000005;
multiset<long long int> se;
multiset<long long int>::iterator it;
map<long long int, long long int> ma;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int mi = inf, maxi = -1, sum, ans = 0;
  long long int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    if (ma.count(a) == 0)
      ma[a] = 1;
    else
      ma[a]++;
    se.insert(a);
    maxi = max(maxi, a);
    sum = 2 * maxi;
  }
  for (it = se.begin(); it != se.end(); it++) {
    long long int val = 1, curr = *it;
    while (val <= sum) {
      if (val - curr > curr && se.find(val - curr) != se.end())
        ans += ma[val - curr];
      else if (val - curr == curr && ma[curr] > 1) {
        ans += ma[curr] - 1;
        ma[curr]--;
      }
      val *= 2;
    }
  }
  cout << ans << endl;
  return 0;
}
