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;
const int N = 100010;
long long n, a[N];
void init() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
}
long long check(long long now) {
  long long ans = 0;
  map<long long, long long> ma;
  map<long long, long long>::iterator it = ma.begin();
  for (long long i = 0; i < n; ++i) {
    if (ma.count(now - a[i]) > 0) {
      ans += ma[now - a[i]];
    }
    ++ma[a[i]];
  }
  return ans;
}
void solve() {
  long long ans = 0, pow = 1;
  for (long long i = 0; i < 63; ++i) {
    if (i != 0) pow = pow << 1;
    ans += check(pow);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int __ = 1;
  while (__--) {
    init();
    solve();
  }
}
