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;
const int nmax = 100000;
int n, v[nmax + 5];
long long ans = 0;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> v[i];
  sort(v + 1, v + n + 1);
  for (int i = 1; i <= n; i++) {
    for (long long P = 0; P <= 35; P++) {
      long long X = (1LL << P), caut = X - v[i];
      if (caut <= 0) continue;
      int r1 = 0, r2 = 0, pas = (1 << 16);
      while (pas) {
        if (r1 + pas <= n and v[r1 + pas] < caut) r1 += pas;
        pas /= 2;
      }
      r1++;
      pas = (1 << 16);
      while (pas) {
        if (r2 + pas <= n and v[r2 + pas] <= caut) r2 += pas;
        pas /= 2;
      }
      if (v[r1] == caut) ans += r2 - r1 + 1;
      if (v[i] == caut) ans--;
    }
  }
  cout << ans / 2;
  return 0;
}
