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;
template <typename Arg1>
void ZZ(const char* name, Arg1&& arg1) {
  cerr << name << " = " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void ZZ(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " = " << arg1;
  ZZ(comma, args...);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  map<long long, long long> m;
  for (long long i = (0); i < (n); i++) {
    long long x;
    cin >> x;
    m[x]++;
  }
  long long ans = 0, same = 0;
  for (auto& [x, y] : m) {
    for (long long i = (0); i < (32); i++) {
      long long req = (1 << i) - x;
      if (req <= 0) continue;
      if (m.count(req) && req != x) ans += y * m[req];
      if (req == x) same += (y * (y - 1)) / 2;
    }
  }
  ans /= 2;
  ans += same;
  cout << ans;
}
