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 N = 1e5 + 9;
long long n, a[N];
double ans;
bool fl;
map<long long, long long> mp;
int main() {
  scanf("%lld", &n);
  for (long long i = 0; i < n; mp[a[i]]++, i++) scanf("%lld", &a[i]);
  for (auto i = mp.begin(); i != mp.end(); i++) {
    for (long long j = 1; j <= 34; j++) {
      map<long long, long long>::iterator it;
      if (i->first <= (1ll << j))
        it = mp.find(-(i->first) + (1ll << j));
      else
        continue;
      if (it != mp.end()) {
        if (it->first == i->first)
          ans += double(i->second * (i->second - 1)) / 2.0;
        else
          ans += double(i->second * it->second) / 2.0;
      }
    }
  }
  printf("%.0f", ans);
}
