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.util.*;
import java.lang.*;
import java.io.*;
import java.util.concurrent.LinkedBlockingQueue;
public class Codechef
{
    static int MAX = (int)2e5+1;
    static int k =0,n=0,m=0;
    static int arr[];
    static ArrayList<Integer>temp;
    static HashMap<Integer,Integer>map = new HashMap<>();
	public static void main (String[] args) throws java.lang.Exception
	{
	    InputReader input = new InputReader(System.in);
	    OutputWriter out = new OutputWriter(System.out);
         n = input.readInt();
         k = input.readInt();
         arr = new int[n];
         for(int i =0;i<n;i++){
             arr[i]= input.readInt();
             map.put(arr[i],map.getOrDefault(arr[i],0)+1);
         }
         int l =0;
         int r = n;
         int res =0;
         while(r-l>1){
             int mid = (l+r)>>1;
             if(check(mid)){
                 l = mid;
             }else{
                 r = mid;
             }
         }
         if(!check(r)){
             check(l);
         }
         for(int val:temp){
             System.out.print(val+" ");
         }
         System.out.println();
	}
	public static boolean check(int val){
	    temp = new ArrayList<>();
	    for(int i =1;i<MAX;i++){
	        if(!map.containsKey(i))continue;
	        int need = Math.min(map.get(i)/val,k-temp.size());
	        for(int j =0;j<need;j++){
	            temp.add(i);
	        }
	    }
	    return temp.size() == k;
	}
}
class Query{
    String s = null;
    public Query(String s){
        this.s = s;
    }
}
class Node{
    int id,r,l;
    public Node(int r,int l,int id){
        this.r = r;
        this.l = l;
        this.id = id;
    }
}
 class InputReader{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		
		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}
		
		public int readInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public String readString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		
		public double readDouble() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') {
				if (c == 'e' || c == 'E') {
					return res * Math.pow(10, readInt());
				}
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') {
				c = read();
				double m = 1;
				while (!isSpaceChar(c)) {
					if (c == 'e' || c == 'E') {
						return res * Math.pow(10, readInt());
					}
					if (c < '0' || c > '9') {
						throw new InputMismatchException();
					}
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
		
		public long readLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public boolean isSpaceChar(int c) {
			if (filter != null) {
				return filter.isSpaceChar(c);
			}
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		
		public String next() {
			return readString();
		}
		
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}
	
 class OutputWriter {
		private final PrintWriter writer;
		
		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}
		
		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}
		
		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0) {
					writer.print(' ');
				}
				writer.print(objects[i]);
			}
			writer.flush();
		}
		
		public void printLine(Object... objects) {
			print(objects);
			writer.println();
			writer.flush();
		}
		
		public void close() {
			writer.close();
		}
		
		public void flush() {
			writer.flush();
		}
	}
	   
//********************************************************************************************************************************
//////////////////////////////////NOTESSS//////////////////////////////////////////////////////////////////////
/*remember substring length you are dealing with take only those substrings
chain like pattern -> linked list (or) to from Arrays
finding degree can be done during input scanning itself(ok&=true:false flag trick)
longest subsquence can be solved in O(n) itself incase of consecutive numbers why try all subsequence go o dp[i-1]
(*******consecutive numberssssss*********)
initialize all global variables and collection ds
swapping changes the arr so make sure initialize with the original again(careful from where to start after swapping)
missing out character when comparing ==0 instead of =='0'
keep going until a condition and start off from first ****stack***
meet in the middle algo
one dfs call will tell u the dfs of all nodes
2d prefix array for max occurence of number in a range.also storing positions in case of duplicate numbers can
help bring down the complexity to o(N) and considering ranges based on positions

when u want to apply changes for a range PREFIX SUMS
want to know number of elements in  range use preifx sum (or)l,r pointer from a particular point
when u want to use binary search check if problem remains the same even after splitting to two parts
checking for range can be done using binary search i.e can this range stisfy a particular condition if so
include more elements to tha range from left to right or right to left depending on the problem
*/