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 long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second < b.second;
}
template <typename T>
void takeInput(vector<T>& v) {
  for (auto& it : v) cin >> it;
}
long long binPow(long long b, long long p) {
  if (p == 0) return 1;
  long long sq = binPow(b, p / 2);
  sq *= sq;
  if (p & 1) sq *= b;
  return sq;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<long long> v(n);
  map<long long, long long> mp;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  long long i = 2;
  vector<long long> power;
  while (i <= 1e10) {
    power.push_back(i);
    i *= 2;
  }
  long long cnt = 0;
  for (int i = 0; i < n; i++) {
    auto it = upper_bound(((power).begin()), ((power).end()), v[i]);
    --mp[v[i]];
    for (; it != power.end(); it++) {
      if (mp[(*it) - v[i]]) {
        cnt += mp[(*it) - v[i]];
      }
    }
  }
  cout << cnt;
  return 0;
}
