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 MOD = 1000 * 1000 * 1000 + 7;
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long MOD2 = 998244353;
const long long N = 1000 * 100 + 10;
const long long N2 = 70;
const long double PI = 3.141592653589793;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p = LLONG_MAX) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  std::vector<long long> p2;
  for (long long i = 0; i < 33; ++i) {
    p2.push_back(1ll << i);
  }
  map<long long, long long> m;
  long long n, ans(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    ans += m[x];
    for (auto y : p2) m[y - x]++;
  }
  cout << ans;
  cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC
       << "ms\n";
  ;
  return 0;
}
