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;
void solve() {
  long long N;
  cin >> N;
  vector<long long> A(N);
  for (long long i = 0; i < N; ++i) cin >> A[i];
  map<long long, long long> cnt;
  for (long long i = 0; i < N; ++i) ++cnt[A[i]];
  set<long long> twos;
  for (long long k = 1; k <= 2e9; k <<= 1LL) twos.insert(k);
  long long ans = 0;
  for (auto it = cnt.begin(); it != cnt.end(); ++it) {
    for (auto tit = twos.begin(); tit != twos.end(); ++tit)
      if (it->first < *tit and cnt.count(*tit - it->first))
        if (it->first * 2 != *tit)
          ans += 1LL * it->second * cnt[*tit - it->first];
  }
  ans /= 2;
  for (auto it = cnt.begin(); it != cnt.end(); ++it)
    if (twos.count(it->first * 2))
      ans += 1LL * cnt[it->first] * (cnt[it->first] - 1) / 2;
  cout << ans << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
