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.math.*;
import java.security.*;
import java.text.*;
import java.time.LocalTime;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;


public class A {
	

	public static void main(String[] args) {
    	FastScanner scan = new FastScanner();
    	//PrintWriter out = new PrintWriter(System.out);
    	
    	int T = 1;
    	T = scan.nextInt();
    		for(int tt=0; tt<T; tt++) {
    			int n = scan.nextInt(), k = scan.nextInt();
    			 a = scan.readArray(n);
    			int [] freq = new int[n+1];
    			ArrayList<Integer> list = new ArrayList<Integer>();
    			for(int i = 0; i<n; i++) {
    				if(freq[a[i]] < k) {
    					freq[a[i]]++;
    					list.add(i);
    				}
    			}
    			int [] ans = new int[n];
    			while(list.size() % k != 0) {
    				list.remove(list.size()-1);
    			}
    			
    			Collections.sort(list, new IntComparator());
    			int currColor = 1;
    			for(int i: list) {
    				ans[i] = currColor++;
    				if(currColor > k) currColor =1 ;
    			}
    			printIntArray(ans);
    		}
	}
	
	static int [] a;
	
	static class IntComparator implements Comparator<Integer> {

		public int compare(Integer o1, Integer o2) {
			return Integer.compare(a[o1], a[o2]);
		}
		
	}
	
	
	public static void printIntArray(int [] a) {
		for(int i: a) System.out.print(i + " ");
		System.out.println();
	}
	
	public static void swapInt(int a, int b) {
		int temp = a;
		a = b;
		b = temp;
	}

	
	  public static void sort(int [] a) {
	    	ArrayList<Integer> b = new ArrayList<>();
	    	for(int i: a) b.add(i);
	    	Collections.sort(b);
	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
	    }
	  
	  public static void reverse(int [] a) {
	    	ArrayList<Integer> b = new ArrayList<>();
	    	for(int i: a) b.add(i);
	    	Collections.reverse(b);
	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
	    }


		
	
	
    static class FastScanner{
    	
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
    	String next() {
    		while(!st.hasMoreTokens())
    			try {
    				st = new StringTokenizer(br.readLine());
    			} catch (IOException e){
    				e.printStackTrace();
    			}
    		return st.nextToken();
    	}
    	
    	int nextInt() {
    		return Integer.parseInt(next());
    	}
    	
    	int [] readArray(int n) {
    		int [] a = new int[n];
    		for(int i=0; i<n ; i++) a[i] = nextInt();
    		return a;
    	}
    	
    	long nextLong() {
    		return Long.parseLong(next());
    	}
    	
    	
    }
    
    
   
}

