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;
unsigned long long a[100001], two[32];
map<unsigned long long, unsigned long long> m;
int main() {
  for (long long i = 0; i < (32); ++i) two[i] = (1LL << i);
  long long(n);
  scanf("%lld", &n);
  for (long long i = 0; i < (n); ++i) scanf("%lld", &(a[i])), ++m[a[i]];
  sort(a, a + n);
  long long ans = 0;
  for (long long i = 0; i < (n); ++i) {
    unsigned long long j = 1;
    for (; j < 32; ++j)
      if (!(a[i] >> j)) break;
    for (; j < 32; ++j) {
      ans += m[two[j] - a[i]];
      if (two[j] - a[i] == a[i]) --ans;
    }
  }
  printf("%lld\n", (ans / 2));
}
