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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.util.PriorityQueue;

public class codeforces {
	public static PrintWriter out;
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	// The next two void functions are shuffle functions on an array.
	// This way, quick sort encounters the worst case scenario with low probability.
	// A reason for doing this is if we don't a primitive data array to an object array.
	// For safety reasons, it would be preferable to use object arrays unless doing so expends a lot of time.
	public static void shuffleLongArray(long[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; i++) {
        	long tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static void shuffleIntArray(int[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; i++) {
        	int tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static void main (String[] args) throws IOException {
    	MyScanner sc = new MyScanner();
    	out = new PrintWriter(new BufferedOutputStream(System.out));
    	// Write solution here.
    	int n = sc.nextInt();
    	long[] best = new long[n];
    	for (int i=0; i < n; i++) {
    		best[i] = sc.nextLong();
    	}
    	int[] val = new int[60]; // array of v_2 values
    	int[] type = new int[n];
    	for (int i=0; i < n; i++) {
    		long y = best[i];
    		int count = 0;
    		while (y % 2L == 0L) {
    			y = y/2L;
    			count++;
    		}
    		val[count]++;
    		type[i] = count;
    	}
    	int feel = -1;
    	int cap = -1;
    	for (int i=0; i < 60; i++) {
    		if (val[i] > cap) {
    			cap = val[i];
    			feel = i;
    		}
    	}
    	out.println(n-cap);
    	for (int i=0; i < n; i++) {
    		if (type[i] != feel) {
    			out.println(best[i]);
    		}
    	}
    	out.close();
	}
	/*
	static class voiceComp implements Comparator<Long> {
		@Override
		public int compare (Long x, Long y) {
			if (x < y) return -1;
			else if (x > y) return 1;
			return 0;
		}
	}
	*/
}