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;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &A) {
  for (long long i = 0; i < A.size(); i++) os << A[i] << " ";
  os << endl;
  return os;
}
template <>
ostream &operator<<(ostream &os, const vector<vector<long long> > &A) {
  long long N = A.size();
  long long M;
  if (N > 0) M = A[0].size();
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < M; j++) os << A[i][j] << " ";
    os << endl;
  }
  return os;
}
struct edge {
  long long from, to, d, c;
  edge(long long _from = 0, long long _to = 0, long long _d = 0,
       long long _c = 0) {
    from = _from;
    to = _to;
    d = _d;
    c = _c;
  }
  bool operator<(const edge &rhs) const {
    return (d == rhs.d) ? (c < rhs.c) : (d < rhs.d);
  }
};
struct aabb {
  long long x1, y1, x2, y2;
  aabb(long long x1, long long y1, long long x2, long long y2)
      : x1(x1), y1(y1), x2(x2), y2(y2) {}
};
struct flow {
  long long to, cap, rev, cost;
  flow(long long to = 0, long long cap = 0, long long rev = 0,
       long long cost = 0)
      : to(to), cap(cap), rev(rev), cost(cost) {}
};
const long long di[4] = {0, -1, 0, 1};
const long long dj[4] = {-1, 0, 1, 0};
const long long ci[5] = {0, 0, -1, 0, 1};
const long long cj[5] = {0, -1, 0, 1, 0};
const long long LINF = LLONG_MAX / 2;
const long long INF = INT_MAX / 2;
const long double PI = acos(-1);
long long pow2(long long n) { return 1LL << n; }
template <typename T, typename U>
bool chmin(T &x, const U &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T, typename U>
bool chmax(T &x, const U &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
struct initializer {
  initializer() { cout << fixed << setprecision(20); }
};
initializer _____;
long long N, M, K, T, Q, H, W;
signed main() {
  cin >> N;
  vector<long long> A(N);
  for (long long i = 0; i < (long long)(N); i++) cin >> A[i];
  vector<long long> B(100, 0);
  vector<long long> C(N);
  for (long long i = 0; i < (long long)(N); i++) {
    long long a = A[i];
    long long tmp = 0;
    while (a % 2 == 0) {
      a /= 2;
      tmp++;
    }
    B[tmp]++;
    C[i] = tmp;
  }
  long long M = 0, j = -1;
  for (long long i = 0; i < (long long)(B.size()); i++) {
    if (M < B[i]) {
      M = B[i];
      j = i;
    }
  }
  vector<long long> ans;
  for (long long i = 0; i < (long long)(N); i++) {
    if (C[i] != j) {
      ans.push_back(A[i]);
    }
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < (long long)(ans.size()); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}