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;
vector<long long int> v;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y / 2;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long int n, i;
  cin >> n;
  long long int a[n];
  unordered_map<long long int, long long int> mp1;
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    for (long long int j = 0; j < v.size(); j++) {
      ans = ans + mp1[v[j] - a[i]];
    }
    mp1[a[i]]++;
  }
  cout << ans;
}
int main() {
  long long int t = 1;
  for (long long int i = 0; i <= 32; i++) {
    long long int x = pow(2, i);
    v.push_back(x);
  }
  while (t--) {
    solve();
  }
  return 0;
}
