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 __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long MOD = 1e+9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
const long long MAXN = 4e+5 + 7;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  sort((a).begin(), (a).end());
  long long ans = 0;
  for (long long power = 1; power < 32; power++) {
    long long reqd = 1LL << power;
    for (long long i = 0; i < n; i++) {
      if (binary_search((a).begin(), (a).end(), reqd - a[i])) {
        if (lower_bound(a.begin() + i + 1, a.end(), reqd - a[i]) != a.end()) {
          long long lb =
              lower_bound(a.begin() + i + 1, a.end(), reqd - a[i]) - a.begin();
          long long ub =
              upper_bound(a.begin() + i + 1, a.end(), reqd - a[i]) - a.begin();
          ans += ub - lb;
        }
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) {
    solve();
  }
}
