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 long long OO = 1LL << 60;
const long double dOO = (double)INT_MAX;
long long mod = 1000000007;
long double eps = 1e-6;
long long a[100009];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long x;
    scanf("%I64d", &x);
    a[i] = x;
  }
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (long long m = 1; m <= 2000000000LL; m <<= 1) {
      long long u = upper_bound(a + i + 1, a + n, m - a[i]) - a;
      long long l = lower_bound(a + i + 1, a + n, m - a[i]) - a;
      ans += u - l;
      if (l == n) break;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
