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.util.*;
  import java.io.*;
  
  import static java.lang.Math.min;
  import static java.lang.Math.max;
  import static java.lang.Math.abs;
  import static java.lang.Math.sqrt;
  
  public class Solution {
  
  	static Scanner scn = new Scanner(System.in);
  	static StringBuilder sb = new StringBuilder();
  	static PrintWriter out = new PrintWriter(System.out);
  
  	// https://randommashupgenerator.netlify.app/
  	// https://a2oj.com/Ladder6.html
  	// https://codeforces.com/problemset?order=BY_RATING_ASC&tags=binary+search%2C1400-1400
  	// https://www.webtoons.com/en/challenge/cp-series/binary-search/viewer?title_no=365151&episode_no=5
  
  	public static void main(String[] HastaLaVistaLa) {
  		int t = scn.nextInt();
  		for(int tests = 0; tests < t; tests++) solve();
  		out.println(sb);
  		out.close();
  	}
  
  	public static void solve() {
  		// Road To Specialist Day 53
  		int n = scn.nextInt();
  		int k = scn.nextInt();
  		int[] a = new int[n];
  		for(int i = 0; i < n; i++) a[i] = scn.nextInt();
  		int[] sorted = new int[n];
  		for(int i = 0; i < n; i++) sorted[i] = a[i];
  		sort(sorted);
  		int[] ans = new int[n];
  		int[] freq = new int[n + 1];
  		int cnt = 1;
  		for(int i = 0; i < n; i++) {
  			int ele = sorted[i];
  			freq[ele]++;
  			if(freq[ele] > k)
  				continue;
  			ans[i] = cnt;
  			cnt++;
  			cnt %= (k + 1);
  			if(cnt == 0) cnt = 1;
  		}
  		int min = Integer.MAX_VALUE;
  		int[] freqK = new int[k + 1];
  		for(int i = 0; i < n; i++) freqK[ans[i]]++;
  		freqK[0] = Integer.MAX_VALUE;
  		for(int i : freqK) {
  			if(i == 0) continue;
  			min = min(min, i);
  		}
  		Arrays.fill(freq, 0);
  		for(int i = 0; i < n; i++) {
  			int ele = ans[i];
  			freq[ele]++;
  			if(freq[ele] > min) ans[i] = 0;
  		}
  		HashMap<Integer, List<Integer>> map = new HashMap<>();
  		for(int i = 0; i < n; i++) {
  			int sot = sorted[i];
  			int answer = ans[i];
  			if(map.containsKey(sot)) {
  				List<Integer> l = map.get(sot);
  				l.add(answer);
  				map.put(sot, l);
  			}else {
  				List<Integer> l = new ArrayList<>();
  				l.add(answer);
  				map.put(sot, l);
  			}
  		}
  		for(int i = 0; i < n; i++) {
  			int ele = a[i];
  			List<Integer> l = map.get(ele);
  			sb.append(l.get(l.size() - 1) + " ");
  			l.remove(l.size() - 1);
  			map.put(ele, l);
  		} 
  		sb.append("\n");
  	}
  	public static void sort(int[] a) {
  		List<Integer> l = new ArrayList<>();
  		for(int i : a) l.add(i);
  		Collections.sort(l);
  		int s = 0;
  		for(int i : l) a[s++] = i;
  	}
  }