This problem is an extension of the problem "Wonderful Coloring - 1". It has quite many differences, so you should read this statement completely.

Recently, Paul and Mary have found a new favorite sequence of integers a_1, a_2, ..., a_n. They want to paint it using pieces of chalk of k colors. The coloring of a sequence is called wonderful if the following conditions are met:

  1. each element of the sequence is either painted in one of k colors or isn't painted; 
  2. each two elements which are painted in the same color are different (i. e. there's no two equal values painted in the same color); 
  3. let's calculate for each of k colors the number of elements painted in the color — all calculated numbers must be equal; 
  4. the total number of painted elements of the sequence is the maximum among all colorings of the sequence which meet the first three conditions. 



E. g. consider a sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. One of the wonderful colorings of the sequence is shown in the figure.

<image> The example of a wonderful coloring of the sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. Note that one of the elements isn't painted.

Help Paul and Mary to find a wonderful coloring of a given sequence a.

Input

The first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases. Then t test cases follow.

Each test case consists of two lines. The first one contains two integers n and k (1 ≤ n ≤ 2⋅10^5, 1 ≤ k ≤ n) — the length of a given sequence and the number of colors, respectively. The second one contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

Output t lines, each of them must contain a description of a wonderful coloring for the corresponding test case.

Each wonderful coloring must be printed as a sequence of n integers c_1, c_2, ..., c_n (0 ≤ c_i ≤ k) separated by spaces where

  * c_i=0, if i-th element isn't painted; 
  * c_i>0, if i-th element is painted in the c_i-th color. 



Remember that you need to maximize the total count of painted elements for the wonderful coloring. If there are multiple solutions, print any one.

Example

Input


6
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9


Output


1 1 0 2 3 2 2 1 3 3
4 2 1 3
1
0 0 1 1 0 1 1 1 0 1 1 1 0
2 1 2 2 1 1 1 1 2 1 0 2 2
1 1 3 2 1 3 3 1 2 2 3 2 0

Note

In the first test case, the answer is shown in the figure in the statement. The red color has number 1, the blue color — 2, the green — 3.
12
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9
 6
10 3
3 1 1 1 1 10 3 10 10 2
4 4
1 1 1 1
1 1
1
13 1
3 1 4 1 5 9 2 6 5 3 5 8 9
13 2
3 1 4 1 5 9 2 6 5 3 5 8 9
13 3
3 1 4 1 5 9 2 6 5 3 5 8 9

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 1 0 
2 1 2 2 1 1 1 1 2 1 0 2 2 
1 1 3 2 1 3 3 1 2 2 3 2 0 

import java.util.*;
  import java.io.*;
  public class A
    {
        static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
        static StringTokenizer st=null;
        static String next()
        {
            while(st==null || !st.hasMoreElements())
            {
                try{
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        static int nextInt()
        {
            return Integer.parseInt(next());
        }
        static long nextLong()
        {
            return Long.parseLong(next());
        }
        
        static double nextDouble(){
            return Double.parseDouble(next());
        }
        
     
       public static void main(String[] args) {
            try{
                
                int t=nextInt();
                for(int i=0;i<t;i++){
                    
                    int n=nextInt();
                    int k=nextInt();
                    
                    int a[]=new int[n];
                    int nums[]=new int[n+1];
                    HashMap<Integer,TreeSet<Integer>> hmpos=new HashMap<>();
                    
                    for(int j=0;j<n;j++){
                        a[j]=nextInt();
                        if(hmpos.get(a[j])==null){
                            TreeSet<Integer> ts=new TreeSet<>();
                            ts.add(j);
                            hmpos.put(a[j], ts);
                        }else{
                            TreeSet<Integer> ts=hmpos.get(a[j]);
                            ts.add(j);
                            hmpos.put(a[j], ts);
                        }
                        nums[a[j]]++;
                    }
                    
                    int ans[]=new int[n];
                    TreeSet<Integer> tsk=new TreeSet<>();
                    
                    for(int j=0;j<n;j++){
                        
                        if(nums[a[j]]>=k){
                            nums[a[j]]=k;
                            tsk.add(a[j]);
                        }
                        
                        if(nums[a[j]]>0 && tsk.contains(a[j])==true){
                            int set=nums[a[j]];
                            ans[j]=set;
                            nums[a[j]]--;
                        }
                    }
                    
                    int remaining=0;
                    TreeSet<Integer> remainingNum=new TreeSet<>();
                    
                    for(int j=0;j<n;j++){
                        if(nums[a[j]]>0 && remainingNum.contains(a[j])==false){
                            
                           remainingNum.add(a[j]);
                           remaining+=nums[a[j]];
                        }
                    }
                    
                    int eachSet=remaining/k;
                    int set=1;
                    int count=1;
                    int totalnum=0;
                    
                    while(!remainingNum.isEmpty()){
                        int num=remainingNum.pollFirst();
                        
                        
                        TreeSet<Integer> posts=hmpos.get(num);
                        while(!posts.isEmpty()){
                            int pos=posts.pollFirst();
                            ans[pos]=set;
                            count++;
                            totalnum++;
                            if(totalnum==(eachSet*k)){
                                break;
                            }
                            
                            if(count>eachSet){
                                set++;
                                count=1;
                            }
                        }
                        hmpos.put(num, posts);
                        
                        if(totalnum==(eachSet*k)){
                            break;
                        }
                        
                    }
                    
                    for(int j=0;j<n;j++){
                        bw.write(ans[j]+" ");
                    }
                    bw.write("\n");
                   
                    
                    
                }
                
    		bw.flush();
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    }  