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.*;
public class Codechef
{       static PrintWriter out=new PrintWriter(System.out);static FastScanner in = new FastScanner(System.in);static class FastScanner {BufferedReader br;StringTokenizer stok;FastScanner(InputStream is) {br = new BufferedReader(new InputStreamReader(is));}
        String next() throws IOException {while (stok == null || !stok.hasMoreTokens()) {String s = br.readLine();if (s == null) {return null;}
                stok = new StringTokenizer(s);}return stok.nextToken();}
        int ni() throws IOException {return Integer.parseInt(next());}long nl() throws IOException {return Long.parseLong(next());}double nd() throws IOException {return Double.parseDouble(next());}char nc() throws IOException {return (char) (br.read());}String ns() throws IOException {return br.readLine();}
        int[] nia(int n) throws IOException{int a[] = new int[n];for (int i = 0; i < n; i++)a[i] = ni();return a;}long[] nla(int n) throws IOException {long a[] = new long[n];for (int i = 0; i < n; i++)a[i] = nl();return a;}
        double[] nda(int n)throws IOException {double a[] = new double[n];for (int i = 0; i < n; i++)a[i] = nd();return a;}int [][] imat(int n,int m) throws IOException{int mat[][]=new int[n][m];for(int i=0;i<n;i++){for(int j=0;j<m;j++)mat[i][j]=ni();}return mat;}
    }
	static long mod=Long.MAX_VALUE;
	static int k=0;
	static HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
	public static void main (String[] args) throws java.lang.Exception
	{   int i,j;
	    ArrayList<Integer> arr=new ArrayList<Integer>();
        HashSet<Integer> set=new HashSet<Integer>();
        PriorityQueue<Integer> pq=new PriorityQueue<Integer>();
        int n=in.ni();
        k=in.ni();
        int a[]=in.nia(n);
        for(int z:a)
        {   if(hm.containsKey(z))
		        hm.put(z,hm.get(z)+1);
		    else
		        hm.put(z,1);
        }
        //out.println(hm);
        int ans=binarySearch(1,100005);
        //out.println(ans);
        for(int c:hm.keySet())
        {   int z=hm.get(c)/ans;
            for(i=1;i<=z;i++)
            {   if(arr.size()==k)
                    break;
                else
                    arr.add(c);
            }
        }
        for(int c:arr)
            out.print(c+" ");
        out.close();
	}
	static int binarySearch(int low, int high) 
    { 
        if (high >= low) 
        {   int mid = (high+low+1)/2;  
            int z=f(mid);
            //out.println("low="+low+" high="+high);
            //out.println("mid="+mid+" f(mid)="+z);
            if (f(mid) >=0 && f(mid+1)<0) 
            {   if(f(mid+1)<0) 
                    return mid; 
                else
                    return mid+1;
            }    
            if (f(mid) < 0) 
                return binarySearch(low,mid-1); 
            else
                return binarySearch(mid,mid+10000); 
            //else
            //    return binarySearch(mid,high);
        } 
        return -1; 
    } 
	static int f(int a)
	{   int count=0;     
	    for(int c:hm.keySet())
	        count+=hm.get(c)/a;
	    return count-k;     
	}
	
	static class pair implements Comparable<pair>{
		int x, y;
		public pair(int x, int y){this.x = x; this.y = y;}
		@Override
		public int compareTo(pair arg0) 
		{   if(x<arg0.x)    return -1;
		    else if(x==arg0.x)
		    {   if(y<arg0.y)    return -1;
		        else if(y>arg0.y)   return 1;
		        else    return 0;
		    }
		    else    return 1;
		}
		public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof pair)) return false;
        pair key = (pair) o;
        return x == key.x && y == key.y;}
        public int hashCode() {int result = x;result = 31 * result + y;return result;}
	}
	static long gcd(long a,long b)
	{   if(b==0)
	        return a;
	    return gcd(b,a%b);    
	}
	static long exponent(long a,long n)
	{   long ans=1;
	    while(n!=0)
	    {   if(n%2==1)
	            ans=(ans*a)%mod;
	       a=(a*a)%mod;
	       n=n>>1;
	    }
	    return ans;
	}
	static void merge(int arr[], int l, int m, int r) {   
        int n1 = m - l + 1; int n2 = r - m; int L[] = new int [n1]; int R[] = new int [n2]; 
        
        for (int i=0; i<n1; ++i) L[i] = arr[l + i]; for (int j=0; j<n2; ++j) R[j] = arr[m + 1+ j]; int i = 0, j = 0; int k = l; while (i < n1 && j < n2) {   if (L[i] <= R[j]) {   arr[k] = L[i]; i++; } else{   arr[k] = R[j]; j++; } k++; } while (i < n1){   arr[k] = L[i]; i++; k++; } while (j < n2) { arr[k] = R[j];   j++;     k++; } 
    } 
    static void Sort(int arr[], int l, int r) {if (l < r) {   int m = (l+r)/2; Sort(arr, l, m); Sort(arr , m+1, r); merge(arr, l, m, r); } } 
    
    static void sort(int a[])
    {Sort(a,0,a.length-1);}
}