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

public class CF_1551b2{
    public static final void main(String[] args){
        Kattio io= new Kattio();
        int t= io.getInt();
        while(t-->0){
            int n= io.getInt(), k= io.getInt();
            int[] a= new int[n];
            int[] f= new int[n+1];
            int[][] use= new int[n+1][];
            int[] idx= new int[n+1];
            for(int i=0; i<n; i++){
                a[i]= io.getInt();
                f[a[i]]++;
            }
            //err("f=%s", Arrays.toString(f));
            int nonFull= 0;                     // need to have same amount of all colors
            for(int v=1; v<=n; v++){
                if(f[v]<k) nonFull+= f[v];      // for those with f[v]>=k we'll always use k colors
            }
            int nonFullColored= k*(nonFull/k);  // for those with f[v]>l we'll allocate colors in loop
            
            //err("n=%d, k=%d, nonFull=%d, nonFullColored=%d", n, k, nonFull, nonFullColored);

            // allocate colors to values
            for(int v=1, used=0, c=1; v<=n; v++){
                if(f[v]>=k){ // -- first k colored
                    use[v]= new int[k+1];
                    for(int i=1; i<=k; i++){
                        use[v][++idx[v]]= i;
                    }
                }else if(f[v]>0 && used<nonFullColored){ //f[v]<k -- nonFullColored
                    use[v]= new int[f[v]+1];
                    for(int i=1; i<=f[v] && used<nonFullColored; i++){
                        use[v][++idx[v]]= c++;
                        if(c>k) c=1;
                        used++;
                    }
                }else if(f[v]>0){
                    use[v]= new int[]{0};
                }
            }   
            // create colored array
            StringBuilder ans= new StringBuilder();
            for(int i=0; i<n; i++){
                int v= a[i];
                int c= use[v][idx[v]];
                if(idx[v]>0) --idx[v];
                ans.append(c);
                if(i!=n-1) ans.append(' ');
            }                                 
            io.println(ans);
        }
        io.close();
    }

    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private String line, token;
        private StringTokenizer st;
    
        public Kattio(){this(System.in);}
        public Kattio(InputStream i){
        super(new BufferedOutputStream(System.out));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public Kattio(InputStream i, OutputStream o){
        super(new BufferedOutputStream(o));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public boolean isLineEnd(){
            return !st.hasMoreTokens();
        }
        public boolean hasMoreTokens(){
            return peekToken()!=null;
        }
        public int getInt(){
            return Integer.parseInt(nextToken());
        }
        public double getDouble(){ 
            return Double.parseDouble(nextToken());
        }
        public long getLong(){
            return Long.parseLong(nextToken());
        }
        public String getWord(){
            return nextToken();
        }
        private String peekToken(){
            if(token==null) try {
                while(st==null || !st.hasMoreTokens()) {
                    line= r.readLine();
                    if(line==null) return null;
                    st= new StringTokenizer(line);
                }
                token= st.nextToken();
            }catch(IOException e){}
            return token;
        }
        private String nextToken() {
            String ans= peekToken();
            token= null;
            return ans;
        }
    }    
}