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;
void adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0},
          dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
long long nod(long long a, long long b) {
  if (b > a) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nok(long long a, long long b) { return a * b / nod(a, b); }
void sp(long long a, double b) { cout << fixed << setprecision(a) << b; }
template <class A>
void read(vector<A>& v);
template <class A, size_t S>
void read(array<A, S>& a);
template <class T>
void read(T& x) {
  cin >> x;
}
void read(double& d) {
  string t;
  read(t);
  d = stod(t);
}
void read(long double& d) {
  string t;
  read(t);
  d = stold(t);
}
template <class H, class... T>
void read(H& h, T&... t) {
  read(h);
  read(t...);
}
template <class A>
void read(vector<A>& x) {
  for (auto& a : x) read(a);
}
template <class A, size_t S>
void read(array<A, S>& x) {
  for (auto& a : x) read(a);
}
string to_string(char c) { return string(1, c); }
string to_string(const char* s) { return string(s); }
string to_string(string s) { return s; }
string to_string(vector<bool> v) {
  string res;
  for (int i = (0); (1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size());
       i += (1))
    res += char('0' + v[i]);
  return res;
}
template <size_t S>
string to_string(bitset<S> b) {
  string res;
  for (int i = (0); (1) > 0 ? i < (S) : i > (S); i += (1))
    res += char('0' + b[i]);
  return res;
}
template <class T>
string to_string(T v) {
  bool f = 1;
  string res;
  for (auto& x : v) {
    if (!f) res += ' ';
    f = 0;
    res += to_string(x);
  }
  return res;
}
template <class A>
void write(A x) {
  cout << to_string(x);
}
template <class H, class... T>
void write(const H& h, const T&... t) {
  write(h);
  write(t...);
}
void print() { write("\n"); }
template <class H, class... T>
void print(const H& h, const T&... t) {
  write(h);
  if (sizeof...(t)) write(' ');
  print(t...);
}
void yes(bool ok) { print((ok ? "YES" : "Yes")); }
void no(bool ok) { print((ok ? "NO" : "No")); }
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
const int M = 1005;
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return ((binpow(a, n - 1) % MOD) * a) % MOD;
  else {
    long long b = (binpow(a, n / 2)) % MOD;
    return (b * b) % MOD;
  }
}
long long cnt[N], n, a[N], ans = 0;
map<long long, long long> m;
void solve() {
  read(n);
  cnt[0] = 1;
  for (int i = 1; i <= 30; ++i) cnt[i] = cnt[i - 1] * 2;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    for (int j = 1; j <= 30; ++j) {
      if (cnt[j] - a[i] >= 0) ans += m[cnt[j] - a[i]];
    }
    m[a[i]]++;
  }
  print(ans);
}
int main() {
  adskiy_razgon();
  long long t = 1;
  for (int i = (0); (1) > 0 ? i < (t) : i > (t); i += (1)) {
    solve();
  }
  return 0;
}
