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 maxn = 100001;
int T, n;
long long a[maxn], ans = 0;
map<int, int> f;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    f[a[i]]++;
  }
  for (int i = 1; i <= 31; i++) {
    long long num = pow(2.0, i * 1.0);
    for (int j = 1; j <= n; j++) {
      long long tmp = num - a[j];
      if (tmp == a[j])
        ans += (f[tmp] - 1);
      else
        ans += f[tmp];
    }
  }
  printf("%I64d", ans / 2);
  return 0;
}
