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>
int pow2[31];
int N;
int val[100005];
void input() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", val + i);
}
long long solve() {
  long long ans(0LL);
  std::sort(val, val + N);
  pow2[0] = 1;
  for (int i = 1; i <= 30; i++) pow2[i] = pow2[i - 1] << 1;
  for (int i = 0; i < N; i++) {
    int f = std::upper_bound(pow2, pow2 + 31, val[i]) - pow2;
    for (int j = f; j <= 30; j++)
      ans += (std::upper_bound(val + i + 1, val + N, pow2[j] - val[i]) -
              std::lower_bound(val + i + 1, val + N, pow2[j] - val[i]));
  }
  return ans;
}
int main() {
  input();
  printf("%I64d\n", solve());
  return 0;
}
