Problem: Boy Dima gave Julian a birthday present — set B consisting of positive integers. However, he didn't know, that Julian hates sets, but enjoys bipartite graphs more than anything else!

Julian was almost upset, but her friend Alex said, that he can build an undirected graph using this set in such a way: let all integer numbers be vertices, then connect any two i and j with an edge if |i - j| belongs to B.

Unfortunately, Julian doesn't like the graph, that was built using B. Alex decided to rectify the situation, so he wants to erase some numbers from B, so that graph built using the new set is bipartite. The difficulty of this task is that the graph, Alex has to work with, has an infinite number of vertices and edges! It is impossible to solve this task alone, so Alex asks you for help. Write a program that erases a subset of minimum size from B so that graph constructed on the new set is bipartite.

Recall, that graph is bipartite if all its vertices can be divided into two disjoint sets such that every edge connects a vertex from different sets.

Input

First line contains an integer n ~ (1 ⩽ n ⩽ 200 000) — size of B

Second line contains n integers b_1, b_2, …, b_n ~ (1 ⩽ b_i ⩽ 10^{18}) — numbers of B, all b_i are unique

Output

In the first line print single integer k – the number of erased elements. In the second line print k integers — values of erased elements.

If there are multiple answers, print any of them.

Examples

Input


3
1 2 3


Output


1
2 

Input


2
2 6


Output


0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void optimise() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int mod = 1e9 + 7;
void normalize(long long int &a) { a = (a + mod) % mod; }
long long int modmul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a * b) % mod;
}
long long int modadd(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a + b) % mod;
}
long long int modsub(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a - b + mod) % mod;
}
long long int me(long long int x, long long int n) {
  x %= mod;
  if (n == 0) return 1;
  long long int u = me(x, n / 2) % mod;
  u = (u * u) % mod;
  if (n % 2) u = (u * x) % mod;
  return u;
}
long long int me1(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int u = me1(x, n / 2);
  u = u * u;
  if (n % 2) u = u * x;
  return u;
}
inline long long int modInv(long long int a) { return me(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  return modmul(a, modInv(b));
}
float power(float x, long long int y) {
  float temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else {
    if (y > 0)
      return x * temp * temp;
    else
      return (temp * temp) / x;
  }
}
long long int __gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return __gcd(b, a % b);
}
pair<long long int, long long int> operator+(
    pair<long long int, long long int> a,
    pair<long long int, long long int> b) {
  return {a.first + b.first, a.second + b.second};
}
pair<long long int, long long int> operator-(
    pair<long long int, long long int> a,
    pair<long long int, long long int> b) {
  return {a.first - b.first, a.second - b.second};
}
pair<long long int, long long int> operator*(
    pair<long long int, long long int> a,
    pair<long long int, long long int> b) {
  return {a.first * b.first, a.second * b.second};
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (auto it : v) os << it << " ";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (long long int i = 0; i < v.size(); ++i) os << v[i] << " ";
  return os;
}
bool compare(long long int a, long long int b) { return abs(a) < abs(b); }
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v[101];
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    long long int k = 0;
    long long int n1 = x;
    while (x % 2 == 0) {
      x /= 2;
      k++;
    }
    v[k].push_back(n1);
  }
  long long int ans = 0;
  long long int max = 0;
  for (long long int i = 0; i <= 100; i++) {
    if (v[i].size() > max) {
      max = v[i].size();
      ans = i;
    }
  }
  cout << n - max << "\n";
  if (max > 0) {
    for (long long int i = 0; i <= 100; i++) {
      if (i != ans) {
        for (auto j : v[i]) {
          cout << j << " ";
        }
      }
    }
  }
}
signed main() {
  optimise();
  long long int t;
  t = 1;
  cout << fixed << setprecision(5);
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "\n"
       << "Time Elasped : " << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
}