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 Aqueous {
		
		
		static MyScanner sc = new MyScanner();
			
		public static void main(String[] args) {
			
			int t = sc.nextInt();
			
				while(t-->0) {
					
					int n = sc.nextInt();
					int k = sc.nextInt();
					
					int a[] = new int[n];
					HashMap<Integer, ArrayList<Integer>> hm = new HashMap<Integer, ArrayList<Integer>>();
					
						for(int i = 0; i<n; i++) {
							a[i] = sc.nextInt();
							
							if(hm.containsKey(a[i])) {
								
								hm.get(a[i]).add(i);
							}
							else {
								hm.put(a[i], new ArrayList<Integer>());
								hm.get(a[i]).add(i);
							}
						}
						
						int res[] = new int[n];
						
						int col = 1;
						
						HashMap<Integer, ArrayList<Integer>> hh = new HashMap<Integer, ArrayList<Integer>>();
						
						for(int i =1; i<=k; i++) {
							hh.put(i, new ArrayList<Integer>());
						}
						
						for(int e: hm.keySet()) {
							
							int cnt = 0;
							
							for(int i = 0; i<hm.get(e).size(); i++) {
								
								if(cnt>=k) {
									break;
								}
								
								res[hm.get(e).get(i)] = col;
								
								hh.get(col).add(hm.get(e).get(i));
								
								col++;
								cnt++;
								
								if(col>k) {
									col=1;
								}
							}
						}
						
						int mn = Integer.MAX_VALUE;
						
						for(int i =1; i<=k; i++) {
							if(hh.get(i).size()<mn) {
								mn = hh.get(i).size();
							}
						}
						
						for(int x : hh.keySet()) {
							
							int r = hh.get(x).size()-mn;
								
								for(int i = 0; i<r; i++) {
									res[hh.get(x).get(i)]=0;
								}
						}
						
						for(int e: res) {
							System.out.print(e+" ");
						}
						System.out.println();
						
				}
		}
		
		static void ruffleSort(int a[]) {

	          int n = a.length;

	          Random r = new Random();

	          for (int i = 0; i < n; i++) {

	              int oi = r.nextInt(n);
	              int temp = a[oi];
	              a[oi] = a[i];
	              a[i] = temp;
	          }

	          Arrays.sort(a);
	      }
		
		
		static void ruffleSort(long a[]) {

	          int n = a.length;

	          Random r = new Random();

	          for (int i = 0; i < n; i++) {
	              int oi = r.nextInt(n);

	              long temp = a[oi];
	              a[oi] = a[i];
	              a[i] = temp;
	          }

	          Arrays.sort(a);
	      }

		
		
	    public static class MyScanner {
	        BufferedReader br;
	        StringTokenizer st;

	        public MyScanner() {
	            br = new BufferedReader(new InputStreamReader(System.in));
	        }

	        String next() {
	            while (st == null || !st.hasMoreElements()) {
	                try {
	                    st = new StringTokenizer(br.readLine());
	                } catch (IOException e) {
	                    e.printStackTrace();
	                }
	            }
	            return st.nextToken();
	        }

	        int nextInt() {
	            return Integer.parseInt(next());
	        }

	        long nextLong() {
	            return Long.parseLong(next());
	        }

	        double nextDouble() {
	            return Double.parseDouble(next());
	        }

	        String nextLine() {
	            String str = "";
	            try {
	                str = br.readLine();
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	            return str;
	        }

	    }
	}
