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 math_codet;

import java.io.*;
import java.util.*;

public class lets_do {
    InputReader in;
    PrintWriter out;
    Helper_class h;
    final long mod = 1000000007;
    public static void main(String[] args) throws java.lang.Exception{
        new lets_do().run();
    }
    void run() throws Exception{
        in=new InputReader();
        out = new PrintWriter(System.out);
        h = new Helper_class();
        int t = 1;
        while(t-->0)
            solve();
        out.flush();
        out.close();
    }
    int MAXN = 200005;
    void solve(){
        int n = h.ni();
        int k = h.ni();
        int[] arr = new int[n];
        int[] freq = new int[MAXN];
        int i = 0;
        for(i = 0; i < n; i++){
            arr[i] = h.ni();
            freq[arr[i]]++;
        }
        int cnt = 0;
        for(i = 0; i < MAXN; i++){
        	if(freq[i] > 0)
        		cnt++;
        }
        Entity[] arr2 = new Entity[cnt];
        //int cnt = 0;
        int q = 0;
        for(i = 0; i < MAXN; i++){
            if(freq[i] > 0){
                //cnt++;
                arr2[q++] = new Entity(freq[i], i);
            }
        }
        //Entity[] arr2 = new Entity[cnt];
        Arrays.sort(freq);
        int[] arr1 = new int[cnt];
        int p = 0;
        for(i = MAXN - 1; i >= 0; i--){
            if(freq[i] > 0){
                arr1[p] = freq[i];
                p++;
            }
        }
        Arrays.sort(arr1);
        //System.out.println(Arrays.toString(arr1));
        Arrays.sort(arr2, com);
        int lo = 1, hi = n;
        while(lo <= hi){
            int mid = (lo + hi) >> 1;
            boolean ok = false;
            int ind = cnt;
            for(i = 0; i < cnt; i++){
                if(arr1[i] >= mid){
                    ind = i;
                    break;
                }
            }
            int len = 0;
            //if(ind >= 0)
            for(i = ind; i < cnt; i++)
                len += arr1[i] / mid;
            if(len >= k)
                lo = mid + 1;
            else
                hi = mid - 1;
        }
        lo--;
        int cnt1 = 0;
        for(i = 0; i < cnt; i++){
            if(arr1[i] >= lo){
                int z = arr1[i] / lo;
                for(int j = 0; j < z && cnt1 < k; j++){
                    h.p(arr2[i].y+" ");
                    cnt1++;
                }
            }
        }
    }
    static final Comparator<Entity> com=new Comparator<Entity>(){
        public int compare(Entity a, Entity b){
            int xx=Integer.compare(a.x,b.x);
            if(xx==0)
                return Integer.compare(a.y,b.y);
            else
                return xx;
        }
    };
    class Entity{
        int x, y;
        Entity(int p, int q){
            x=p;
            y=q;
        }
    }
    class Helper_class{
        long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
        int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
        int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
        void p(Object o){out.print(o);}
        void pn(Object o){out.println(o);}
        void pni(Object o){out.println(o);out.flush();}
        String n(){return in.next();}
        String nln(){return in.nextLine();}
        int ni(){return Integer.parseInt(in.next());}
        long nl(){return Long.parseLong(in.next());}
        double nd(){return Double.parseDouble(in.next());}

        long mul(long a,long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            a*=b;
            if(a>=mod)a%=mod;
            return a;
        }
        long modPow(long a, long p){
            long o = 1;
            while(p>0){
                if((p&1)==1)o = mul(o,a);
                a = mul(a,a);
                p>>=1;
            }
            return o;
        }
        long add(long a, long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            if(b<0)b+=mod;
            a+=b;
            if(a>=mod)a-=mod;
            return a;
        }
    }

    class InputReader{
        BufferedReader br;
        StringTokenizer st;
        public InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public InputReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

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

        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
    }
}