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(int test_number);
void pre() {
  cout.setf(ios::fixed);
  cout.precision(20);
  cerr.setf(ios::fixed);
  cerr.precision(3);
}
void post() {
  fprintf(stderr, "\n");
  fprintf(stderr, "Execution time: %Lf", (long double)clock());
}
const int MAXN = 100100;
int a[999999];
map<int, int> cnt;
inline void solve(int test_number) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  long long ans = 0LL;
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    for (long long j = 1; j < 2 * (int)(1e+9); j *= 2LL) {
      if (j < a[i]) continue;
      ans += cnt[j - a[i]];
    }
    cnt[a[i]]++;
  }
  printf("%lld", ans);
  return;
}
int main() {
  pre();
  int n = 1;
  for (int i = 0; i < n; i++) {
    solve(i + 1);
  }
  post();
  return 0;
}
