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;
int n;
unordered_map<long long, int> h;
int main() {
  long long a, ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a);
    for (int i = 0; i < 31; i++) {
      long long p = (long long)(1) << i;
      if (p > a && (h.find(p - a) != h.end())) {
        ans += h[p - a];
      }
    }
    if (h.find(a) == h.end()) {
      h[a] = 1;
    } else {
      h[a]++;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
