Problem: You are given an array s consisting of n integers.

You have to find any array t of length k such that you can cut out maximum number of copies of array t from array s.

Cutting out the copy of t means that for each element t_i of array t you have to find t_i in s and remove it from s. If for some t_i you cannot find such element in s, then you cannot cut out one more copy of t. The both arrays can contain duplicate elements.

For example, if s = [1, 2, 3, 2, 4, 3, 1] and k = 3 then one of the possible answers is t = [1, 2, 3]. This array t can be cut out 2 times. 

  * To cut out the first copy of t you can use the elements [1, \underline{2}, 3, 2, 4, \underline{3}, \underline{1}] (use the highlighted elements). After cutting out the first copy of t the array s can look like [1, 3, 2, 4]. 
  * To cut out the second copy of t you can use the elements [\underline{1}, \underline{3}, \underline{2}, 4]. After cutting out the second copy of t the array s will be [4]. 



Your task is to find such array t that you can cut out the copy of t from s maximum number of times. If there are multiple answers, you may choose any of them.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in s and the desired number of elements in t, respectively.

The second line of the input contains exactly n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 2 ⋅ 10^5).

Output

Print k integers — the elements of array t such that you can cut out maximum possible number of copies of this array from s. If there are multiple answers, print any of them. The required array t can contain duplicate elements. All the elements of t (t_1, t_2, ..., t_k) should satisfy the following condition: 1 ≤ t_i ≤ 2 ⋅ 10^5.

Examples

Input


7 3
1 2 3 2 4 3 1


Output


1 2 3 


Input


10 4
1 3 1 3 10 3 7 7 12 3


Output


7 3 1 3


Input


15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1


Output


1 1 

Note

The first example is described in the problem statement.

In the second example the only answer is [7, 3, 1, 3] and any its permutations. It can be shown that you cannot choose any other array such that the maximum number of copies you can cut out would be equal to 2.

In the third example the array t can be cut out 5 times.

Solution (in JAVA):

Solution:
//package Codeforces;
import java.util.*;
import java.io.*;
public class C
{
   static final int mod = (int) 1e9 + 7;
   static final int inf = (int) 1e9;
   static final long INF = (long) 1e18;
   static HashMap<Integer, Integer> map = new HashMap<>();
   static ArrayList<Pair> cnt = new ArrayList<Pair>();
   static int ans[]=new int[(int)2e5+1];static int k;
   public static void main(String[] args) throws IOException 
   {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n=fs.nextInt();k=fs.nextInt();
		int arr[]=new int[n];
		ans=new int[k];
		for(int i=0;i<n;i++)
		{
			arr[i]=fs.nextInt();
			map.put(arr[i],map.getOrDefault(arr[i],0)+1);
		}
		for(int i:map.keySet())
		{
			cnt.add(new Pair(i,map.get(i)));
		}
		Collections.sort(cnt,new Sort());
		int left=0,right=inf;
		while(left<=right)
		{
			int mid=(left+right)/2;
			if(isValid(mid))
			{
			    left=mid+1;
			}
			else
				right=mid-1;
		}
		for(int i=0;i<k;i++)out.print(ans[i]+" ");
		out.close();
   }
   static class Sort implements Comparator<Pair>
   {
	   public int compare(Pair s1,Pair s2)
	   {
		   if(s1.second==s2.second)return 0;
		   else if(s1.second>s2.second)return -1;
		   else return 1;
	   }
   }
   public static boolean isValid(int mid)
   { 
	   if(mid==0)return true;
	   long sum=0;
	   ArrayList<Pair> list = new ArrayList<Pair>();
	   for(int i=0;i<cnt.size();i++)
	   {
		   sum=sum+(cnt.get(i).second/mid);
		   list.add(new Pair(cnt.get(i).second/mid,cnt.get(i).first));
	   }
	   if(sum<k)return false;
	   int index=0;int copy=k;int index2=0;
	   while(copy!=0)
	   {
		   if(list.get(index2).first==0)
		   {
			   index2++;
		   }
		   ans[index]=list.get(index2).second;
		   list.get(index2).first--;
		   copy--;index++;
	   }
	   return true;
   }
   static class Pair
   {
	   int first,second;
	   Pair(int first,int second)
	   {
		   this.first=first;this.second=second;
	   }
   }
	public static int upper_bound(int arr[],int key)
	{
		int index=inf,num=0;
		int left=0,right=arr.length-1;
		while(left<=right)
		{
			int mid=left+(right-left)/2;
			if(arr[mid]<=key) {index=mid;num=arr[mid];left=mid+1;}
			else 
				right=mid-1;
		}
		return index==inf ? arr.length:index;
	}
}
class FastScanner {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = new StringTokenizer("");

	public String next() {
		while (!st.hasMoreTokens())
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		return st.nextToken();

	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	int[] sort(int arr[]) {
		ArrayList<Integer> list = new ArrayList<Integer>();
		for (int i : arr)
			list.add(i);
		Collections.sort(list);
		for (int i = 0; i < arr.length; i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

	char[] charsort(char arr[]) {
		ArrayList<Character> list = new ArrayList<>();
		for (char c : arr)
			list.add(c);
		Collections.sort(list);
		for (int i = 0; i < list.size(); i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

	long[] longsort(long arr[]) {
		ArrayList<Long> list = new ArrayList<Long>();
		for (long i : arr)
			list.add(i);
		Collections.sort(list);
		for (int i = 0; i < arr.length; i++) {
			arr[i] = list.get(i);
		}
		return arr;
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public int[] readArray(int n) {
		int[] arr = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = nextInt();
		return arr;
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public long[] longreadArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}