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 mod = (int)1e9 + 7;
const int mod2 = 998244353;
long long exp(long long taban, long long us, long long md) {
  long long carpan = taban % md;
  if (carpan == 0) return 0;
  long long temp = us;
  long long res = 1;
  while (temp) {
    if (temp % 2) res = (res * carpan) % md;
    temp /= 2;
    carpan = (carpan * carpan) % md;
  }
  return res;
}
long long ebob(long long a, long long b) {
  if (!a) return b;
  return ebob(b % a, a);
}
long long ekok(long long a, long long b) { return (a * b) / ebob(a, b); }
long long mul(long long a, long long b, long long md) { return a * b % md; }
vector<long long> fact;
vector<long long> inv_fact;
void fact_init(int n) {
  fact.resize(n + 5);
  inv_fact.resize(n + 5);
  fact[0] = inv_fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv_fact[i] = exp(fact[i], mod - 2, mod);
  }
}
long long komb(long long a, long long b) {
  if (a < b) return 0;
  return fact[a] * (inv_fact[a - b] * inv_fact[b] % mod) % mod;
}
const int N = 2e5 + 5;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  unordered_map<int, int> mp;
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int s;
    cin >> s;
    for (int u = 0; u < 32; u++) {
      ans += mp[(1 << u) - s];
    }
    mp[s]++;
  }
  cout << ans << endl;
}
