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.InputStream;
import java.io.InputStreamReader;
import java.io.FileReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Main {
    static Scanner scanner = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static HashMap<Integer,Integer> hm;
    public static void main(String[] args) throws Exception {
        //============================================//
        //answer here:
        /*problem CF-div3- 521 D.cutting edge
        the aim of the problem is to find the maximum number of cuts you can make in a given array using
        a subsequence from the given array every time you cut from the array you can only cut the same subsequence
        however the sequence is not necessarily placed in the same order
        */
        int n=scanner.nextInt(), k=scanner.nextInt(); //n is the length of the array, k is the length of the subsequence
        hm=new HashMap<>();
        int[] arr=new int[n];
        for (int i=0; i<n; i++){ //taking the input into array and hashmap recording occurences of each element
            arr[i]=scanner.nextInt();
            hm.put(arr[i],hm.getOrDefault(arr[i],0)+1); //smart way to add to hashmap.
        }
        //binary search for the max number of cuts we can make:
        int start=1; //least number of cuts is 1.
        int end=n; //max number of cuts is n.
        int sol=0;
        while (start<=end){ //monotone function
            int mid=(start+end)>>1;//dividing using arithmetic shift
            if (f(mid,k)) {
                sol=mid;
                start=mid+1;
            }else end=mid-1;
        }
        l1: for (int key: hm.keySet()){
            int val=hm.get(key);
            while (val>=sol){ //keep printing the element as long as its occurences are  more than best.
                out.print(key+" ");
                k--;
                val-=sol;
                if (k==0) break l1;
            }
        }

        //============================================//
        out.close();
    }
    static boolean f(int freq,int n){
        int c=0;
        for (int key:hm.keySet()){
            int val=hm.get(key);
            c+=val/freq;
        }
        return (c>=n); //if no. of elements that can be cut freq 
    }
}

class Scanner {
    StringTokenizer st;
    BufferedReader br;

    public Scanner(InputStream system) {
        br = new BufferedReader(new InputStreamReader(system));
    }

    public Scanner(String file) throws Exception {
        br = new BufferedReader(new FileReader(file));
    }

    public String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

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

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

    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public char nextChar() throws IOException {
        return next().charAt(0);
    }

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

    public boolean ready() throws IOException {
        return br.ready();
    }

    public void waitForInput() throws InterruptedException {
        Thread.sleep(3000);
    }

    int[] readArray(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    long[] readLongArray(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}
class pair implements Comparable<pair>{
    int x;
    int y;

    public pair(int x, int y) {
        this.x = x;
        this.y = y;
    }

    @Override
    public String toString() {
        return "";
    }
    @Override
    public int compareTo(pair o){
        return o.y-y;
    }
}