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 n;
int a[10 * 10000 + 10];
map<int, int> m;
int odd[10 * 10000 + 10];
int even[10 * 10000 + 10];
int main() {
  long long int cnt = 0;
  while (scanf("%d", &n) != EOF) {
    cnt = 0;
    int lena = 1;
    int lenb = 1;
    m.clear();
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) {
      if (a[i] & 1)
        odd[lena++] = a[i];
      else
        even[lenb++] = a[i];
    }
    for (int i = 1; i < lena; i++) {
      if (m[odd[i]]) cnt += m[odd[i]];
      for (int k = 1; k <= 30; k++) m[pow(2, k) - odd[i]]++;
    }
    for (int i = 1; i < lenb; i++) {
      if (m[even[i]]) cnt += m[even[i]];
      for (int k = 1; k <= 30; k++) m[pow(2, k) - even[i]]++;
    }
    printf("%lld\n", cnt);
  }
  return 0;
}
