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 dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int xChange[] = {0, 1, 0, -1};
int yChange[] = {1, 0, -1, 0};
const int N = 100001;
int n, a[N];
long long power[31], ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  power[0] = 1;
  for (int i = 1; i <= 30; i++) power[i] = power[i - 1] * 2;
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 30; j++) {
      long long dif = abs(a[i] - power[j]);
      int up, lo;
      up = upper_bound(a + i + 1, a + n, dif) - a;
      lo = lower_bound(a + i + 1, a + n, dif) - a;
      ans += (up - lo);
    }
  }
  cout << ans;
  return 0;
}
