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: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  public class P3 {
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc = new FastReader();
  		int t;
  		t = sc.nextInt();
  		StringBuilder st = new StringBuilder();
  		while (t-- != 0) {
  			int n=sc.nextInt();
  			int k=sc.nextInt();
  			int a[]=new int[n];
  			int c[]=new int[n];
  			Arrays.fill(c, 0);
  			for(int i=0;i<n;i++) {
  				a[i]=sc.nextInt();
  				c[a[i]-1]++;
  			}
  			int x=0;
  			int co=0;
  			for(int i=0;i<n;i++) {
  				if(c[i]>=k) {
  					x++;
  				}else {
  					co+=c[i];
  				}
  			}
  			int per=x+co/k;
  			if(per==0) {
  				for(int i=0;i<n;i++) {
  					st.append("0 ");
  				}
  				st.append("\n");
  				continue;
  			}
  			ArrayList<ArrayDeque<Integer>> list=new ArrayList<>();
  			for(int i=0;i<n;i++) {
  				list.add(new ArrayDeque<>());
  			}
  			int curr=1;
  			int token=0;
  			for(int i=0;i<n;i++) {
  				if(token>=per*k) {
  					break;
  				}
  				for(int j=0;j<Math.min(k, c[i]);j++) {
  					if(token>=per*k) {
  						break;
  					}
  					list.get(i).add(curr);
  					curr++;
  					if(curr==k+1) {
  						curr=1;
  					}
  					token++;
  				}
  			}
  			for(int i=0;i<n;i++) {
  				if(list.get(a[i]-1).size()==0) {
  					st.append("0 ");
  				}else {
  					st.append(list.get(a[i]-1).getFirst()+" ");
  					list.get(a[i]-1).removeFirst();
  				}
  			}
  			st.append("\n");
  		}
  		System.out.println(st);
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			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;
  		}
  	}
  }
  