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;
long long INF = 1 << 28;
const double PI = acos(-1.0);
int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
int knight[8][2] = {1, 2, 1, -2, 2, 1, 2, -1, -1, 2, -1, -2, -2, 1, -2, -1};
const long double EPS = 1e-7;
const int mx = 300050;
int arr[mx];
map<long long, int> mp;
int main() {
  int n;
  long long x;
  scanf("%d", &n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = n - 1; i >= 0; i--) {
    long long x = 1;
    long long y = arr[i];
    for (int j = 32; j >= 1; j--) {
      x = 1;
      x <<= j;
      if (x < y) break;
      long long z = x - y;
      if (mp.find(z) != mp.end()) {
        ans += mp[z];
      }
    }
    mp[y]++;
  }
  cout << ans << endl;
  return 0;
}
