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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.Comparator;

import java.util.List;
public class Main {

	public static void main(String[] args) throws IOException {
		// TODO 自动生成的方法存根
		StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		in.nextToken();
		int n=(int)in.nval;in.nextToken();
		int k=(int)in.nval;
		int a[]=new int[n];
		int map2[]=new int[200005];
		for(int i=0;i<n;i++)
		{
			in.nextToken();
			a[i]=(int)in.nval;
			map2[a[i]]++;
		}
		List<node>list=new ArrayList<>();
		for(int i=0;i<200005;i++)
		{
			if(map2[i]>0)
			{
			node no=new node(i, map2[i]);//这个点的次数
			list.add(no);
			}
		}
		list.sort(com);
      
		int max=list.get(0).value;//出现最多的次数//不可能超过的次数
		int value=0;//最终出现的次数
		int mid=0;//二分查找
		  /*
         * 二分
         */
	
//		int l=1;int r=max+1;
//		while(l<r)
//		{
//			mid=(l+r)/2;
//			
//			int count=0;
//			for(int i=0;i<list.size();i++)
//			{
//				count+=list.get(i).value/mid;
//				if(count>=k)
//				{
//					value=mid;
//                    l=mid+1;
//					break;
//				}		
//			}
//			if(count<k)
//			{
//				r=mid;
//			}			
//		}
//		mid=value;
		for(int q=1;q<=max;q++)
		{
			//System.out.println(q);
			int count=0;
			for(int i=0;i<list.size();i++)
			{
				count+=list.get(i).value/q;
				if(count>=k)
				{
					value=i;
                    mid=q;//次数
					break;
				}			
			}
			if(count<k)
			{break;}
		}
		int time=0;//次数，到K次就停止
		for(int i=0;i<list.size();i++)
		{
			node team=list.get(i);
			int va=team.value;
			while(va-mid>=0)
			{
				va-=mid;
				time++;
				if(time>k) {break;}
				
				out.print(team.index+" ");
			}
			if(time>k)break;
		}
		out.flush();		
	}
	static Comparator<node>com=new Comparator<node>() {
		@Override
		public int compare(node o1, node o2) {
			// TODO 自动生成的方法存根
			return o2.value-o1.value;
		}		
	};
	static class node
	{
		int index;//数值
		int value;//
		
		public node() {}
		public node(int index,int value)
		{
			this.index=index;
			this.value=value;
		}
	}

}