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 B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
		for(int tt=0;tt<t;tt++)
		{
			int n = fs.nextInt(); int k = fs.nextInt();
			
			int[] arr = fs.readArray(n);
			
			int max = 0;
			for(int i:arr)max = Math.max(max,i);
			
			List<Integer>[]adj = new ArrayList[max+1];
			for(int i=0;i<=max;i++)adj[i] = new ArrayList();
			int[] vis = new int[max+1];
			for(int i=0;i<n;i++)
			{
				adj[arr[i]].add(i);
			}
			
			int[] ans = new int[n];
			
			int size = 0;
			
			for(int i=0;i<n;i++)
			{
				if(vis[arr[i]] == 0) {
				if(adj[arr[i]].size()>k)
				{
					size += k;
				}
				else size += adj[arr[i]].size();
				vis[arr[i]] = 1;
//				size += (Math.min(k,adj[arr[i]].size()));
				}
			}
//			System.out.println(size);
			size/=k;
			int[] maxSize = new int[max+1];
			for(int i=0;i<=max;i++)
			{
				if(adj[i].size()<k)maxSize[i] = adj[i].size();
				else maxSize[i] = k*((adj[i].size())/size);
			}
			adj = new ArrayList[max+1];
			
			for(int i=0;i<=max;i++)adj[i] = new ArrayList();
//			System.out.println(size);
			for(int i=0;i<n;i++)
			{
				if(adj[arr[i]].size()==maxSize[arr[i]])continue;
				adj[arr[i]].add(i);
			}
//			System.out.println(size);
			int start = 1;
			int done = 0;
//			System.out.println(adj[1]);
			me:for(int i=0;i<=max;i++)
			{
				if(done>=size)break;
				for(int ele:adj[i])
				{
					if(start == k)
					{
//						System.out.println("Here");
						done++;
						ans[ele] = start;
						if(done>=size)break me;
						start = 1;
					}
					else
					{
//						System.out.println("here");
						ans[ele] = start;
						start++;
					}
				}
			}
			
			for(int i:ans)
			{
				out.print(i+" ");
			}
			out.println();
			
		}
		out.close();
	}
	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());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp);
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
	public static String rev(String str)
	{
		char[] arr = str.toCharArray();
		char[] ret = new char[arr.length];
		int start = arr.length-1;
		for(char i:arr)ret[start--] = i;
		String ret1 = "";
		for(char ch:ret)ret1+=ch;
		return ret1;
	}
}