description: 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.
solution: import java.io.*;
  import java.util.*;
  import java.util.function.IntConsumer;
  
  
  /*
  
  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
  
  1
  3 1
  1 2 3
   */
  @SuppressWarnings("unchecked")
  
  public class ws2 {
  	
  	static boolean usaco = false;
  	static String file = "";
  	static BufferedReader in;
  	static PrintWriter out;
  	public static void main(String[] args) throws Exception {
  //		usaco = true;
  
  		if(usaco) {
  			in = new BufferedReader(new FileReader(file+".in"));
  			out = new PrintWriter(new FileWriter(file+".out"));	
  		}else {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(new PrintStream(System.out));
  		}
  		int N = pi();
  		for(int i = 0; i<N; i++) {
  			int[] d = pl();
  			int n = d[0];
  			int k = d[1];
  			int[][] nums = new int[n][3];
  			int[] num = pl();
  			for(int j = 0; j<num.length; j++) {
  				nums[j][0] = num[j];
  				nums[j][1] = j;
  			}
  			Arrays.sort(nums,(a,b)->a[0]-b[0]);
  			int last = -1;
  			int start = 0;
  			nums[0][2] = 1;
  			if(k==1)last = 0;
  			int cur = 2<=k?2:1;
  			for(int j = 1; j<nums.length; j++) {
  				if(nums[j][0]==nums[j-1][0]) {
  					if(j-start<k) {
  						nums[j][2] = cur;
  						cur++;
  						
  					}
  				}else {
  					start = j;
  					nums[j][2] = cur;
  					cur++;
  				}
  
  				if(cur>k) {
  					last = j;
  					cur = 1;
  				}
  			}
  			for(int na = nums.length-1; na>last; na--) {
  				nums[na][2] = 0;
  			}
  //			System.out.println(last+" "+Arrays.deepToString(nums));
  			int[] res = new int[n];
  			for(int[] ar: nums) {
  				res[ar[1]] = ar[2];
  			}
  			for(int j =0; j<res.length; j++) {
  				out.print(j==0?res[j]:" "+res[j]);
  			}
  			out.println();
  		}
  		in.close();
  		out.close();
  	}
  	public static long[] pll() throws IOException {
  		return pll(in.readLine());
  	}
  	public static long[] pll(String s) throws IOException {
  		return Arrays.stream(s.split(" ")).mapToLong(Long::valueOf).toArray();
  	}
  	public static int[] pl() throws IOException {
  		return pl(in.readLine());
  	}
  	public static int[] pl(String s) throws IOException {
  		return Arrays.stream(s.split(" ")).mapToInt(Integer::valueOf).toArray();
  	}
  	public static double[] pld() throws IOException {
  		return pld(in.readLine());
  	}
  	public static double[] pld(String s) throws IOException {
  		return Arrays.stream(s.split(" ")).mapToDouble(Double::valueOf).toArray();
  	}
  	static void t(int n, IntConsumer consumer) {
  	    for (int i = 0; i < n; i++) {
  	        consumer.accept(i);
  	    }
  	}
  	public static int pi() throws Exception {
  		return Integer.valueOf(in.readLine());
  	}
  }
  