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 main() {
  ios::sync_with_stdio(false);
  long long int a[100000];
  long long int twos[] = {
      2,        4,        8,         16,        32,        64,
      128,      256,      512,       1024,      2048,      4096,
      8192,     16384,    32768,     65536,     131072,    262144,
      524288,   1048576,  2097152,   4194304,   8388608,   16777216,
      33554432, 67108864, 134217728, 268435456, 536870912, 1073741824};
  map<long long, long long> hash;
  int n;
  unsigned long long int ans = 0;
  int check;
  int cur_length = 1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    ans += hash[a[i]];
    for (int j = 0; j <= 29; j++) {
      if (twos[j] >= a[i]) {
        if (twos[j] == a[i]) {
        }
        check = j;
        break;
      }
    }
    for (long long int j = check; j <= 29; j++) {
      if (9) {
        if (twos[j] - a[i] < 0) {
        }
      }
      hash[twos[j] - a[i]]++;
    }
  }
  cout << ans;
  return 0;
}
