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.*;    //PrintWriter
import java.math.*;  //BigInteger, BigDecimal
import java.util.*;  //StringTokenizer, ArrayList


public class R734_Div3_C
{	
	FastReader in;
	PrintWriter out;
	
	public static void main(String[] args)  {
		new R734_Div3_C().run();
	}
	
	void run()
	{		
		in = new FastReader(System.in);
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}
	
	void solve()
	{
		int t = in.nextInt();
		for (int T = 0; T < t; T++) {
			int n = in.nextInt();
			int k = in.nextInt();
			
			int[] a = new int[n];
			int[] c = new int[n];
			HashMap<Integer,Integer> hm = new HashMap<>();
			
			int cnt = 0;
			ArrayList<Integer> u = new ArrayList<>();
			for (int i = 0; i < n; i++) {				
				a[i] = in.nextInt();
				if (hm.containsKey(a[i])) {
					cnt = hm.get(a[i]) + 1;	
				} else {
					cnt = 1;
					u.add(a[i]);
				}
				hm.put(a[i], cnt);
			}
			
			int tot = 0, empty = 0;
			for (int x : u) {
				if (hm.get(x) > k) {
					empty += (hm.get(x) - k);
				}
			}
			tot = n - empty;
			tot = (tot / k) * k;
			//out.println(tot);
			HashMap<Integer,Integer> hm2 = new HashMap<>();
			int color = 1, placed = 0;
			for (int i = 0; i < n; i++) {
				if (hm2.containsKey(a[i])) {
					cnt = hm2.get(a[i]) + 1;	
				} else {
					cnt = 1;
				}
				hm2.put(a[i], cnt);
				
				if (cnt <= k && placed < tot) {
					c[i] = color;
					placed++;
					color++;
					if (color > k) color = 1;
				}
				out.print(c[i] + " ");					
			}
			out.println();
		}
	}

	//-----------------------------------------------------
	void runWithFiles() {
		in = new FastReader(new File("input.txt"));
		try {
			out = new PrintWriter(new File("output.txt"));
		} 
		catch (FileNotFoundException e) {
			e.printStackTrace();
		}
		solve();
		out.close();
	}
	
	class FastReader
	{
	    BufferedReader br;
	    StringTokenizer tokenizer;
	    
	    public FastReader(InputStream stream)
	    {
	        br = new BufferedReader(new InputStreamReader(stream));
	        tokenizer = null;
	    }
		public FastReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
				tokenizer = null;
			}
			catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}
	    
	    private String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens())
	            try {
	            	tokenizer = new StringTokenizer(br.readLine());
	            }
	            catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        return tokenizer.nextToken();
	    }
		public String nextLine() {
			try	{
				return br.readLine();
			}
			catch(Exception e) {
				throw(new RuntimeException());
			}
		}

	    int nextInt() {
	        return Integer.parseInt(next());
	    }
	    long nextLong() {
	        return Long.parseLong(next());
	    }
	    double nextDouble() {
	        return Double.parseDouble(next());
	    }	    
	    BigInteger nextBigInteger() {
	        return new BigInteger(next());
	    }
	    BigDecimal nextBigDecimal() {
	        return new BigDecimal(next());
	    }
	}
}
