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 java.math.*;
  import java.awt.geom.*;
  
  import static java.lang.Math.*;
  
  public class Solution implements Runnable {
  
  	long mod1 = (long) 1e9 + 7;
  	int mod2 = 998244353;
  
  	public void solve() throws Exception {
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n=sc.nextInt(), k=sc.nextInt();
  			int res[] = new int[n];
  			int temp=0;
  			int arr[] = sc.readArray(n);
  			int freq[] = new int[n+1];
  			HashMap<Integer,ArrayList<Integer>> map=new HashMap<>();
  			for(int i=0;i<n;i++) {
  				freq[arr[i]]++;
  				if(map.containsKey(arr[i])) {
  					map.get(arr[i]).add(i);
  				}
  				else {
  					map.put(arr[i],new ArrayList<>());
  					map.get(arr[i]).add(i);
  				}
  			}
  			Deque<Integer> tempo=new LinkedList<>();
  			for(int i:map.keySet()) {
  				if(map.get(i).size()>=k) {
  					int count=1;
  					for(int j:map.get(i)) {
  						res[j]=count++;
  						if(count==k+1) break; 
  					}
  				}
  				else {
  					for(int j:map.get(i)) {
  						tempo.addLast(j);
  					}
  					int count=1;
  					if(tempo.size()>=k) {
  						while(count<=k) {
  							res[tempo.pollFirst()]=count++;
  						}
  					}
  				}
  			}
  			
  			for(int i:res) {
  				out.print(i+" ");
  			}
  			out.println();
  
  		}
  
  	}
  
  	static long gcd(long a, long b) {
  		if (a == 0)
  			return b;
  		return gcd(b % a, a);
  	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l = new ArrayList<>();
  		for (int i : a)
  			l.add(i);
  		Collections.sort(l);
  		for (int i = 0; i < a.length; i++)
  			a[i] = l.get(i);
  	}
  
  	static long ncr(int n, int r, long p) {
  		if (r > n)
  			return 0l;
  		if (r > n - r)
  			r = n - r;
  
  		long C[] = new long[r + 1];
  
  		C[0] = 1;
  
  		for (int i = 1; i <= n; i++) {
  
  			for (int j = Math.min(i, r); j > 0; j--)
  				C[j] = (C[j] + C[j - 1]) % p;
  		}
  		return C[r] % p;
  	}
  
  	void sieveOfEratosthenes(boolean prime[], int size) {
  		for (int i = 0; i < size; i++)
  			prime[i] = true;
  
  		for (int p = 2; p * p < size; p++) {
  			if (prime[p] == true) {
  				for (int i = p * p; i < size; i += p)
  					prime[i] = false;
  			}
  		}
  	}
  
  	static int LowerBound(int a[], int x) { // smallest index having value >= x
  		int l = -1, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] >= x)
  				r = m;
  			else
  				l = m;
  		}
  		return r;
  	}
  
  	static int UpperBound(int a[], int x) {// biggest index having value <= x
  		int l = -1, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] <= x)
  				l = m;
  			else
  				r = m;
  		}
  		return l + 1;
  	}
  
  	public long power(long x, long y, long p) {
  		long res = 1;
  		// out.println(x+" "+y);
  		x = x % p;
  		if (x == 0)
  			return 0;
  
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y = y >> 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	static Throwable uncaught;
  
  	BufferedReader in;
  	FastScanner sc;
  	PrintWriter out;
  
  	@Override
  	public void run() {
  		try {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(System.out);
  			sc = new FastScanner(in);
  			solve();
  		} catch (Throwable uncaught) {
  			Solution.uncaught = uncaught;
  		} finally {
  			out.close();
  		}
  	}
  
  	public static void main(String[] args) throws Throwable {
  		Thread thread = new Thread(null, new Solution(), "", (1 << 26));
  		thread.start();
  		thread.join();
  		if (Solution.uncaught != null) {
  			throw Solution.uncaught;
  		}
  	}
  
  }
  
  class FastScanner {
  
  	BufferedReader in;
  	StringTokenizer st;
  
  	public FastScanner(BufferedReader in) {
  		this.in = in;
  	}
  
  	public String nextToken() throws Exception {
  		while (st == null || !st.hasMoreTokens()) {
  			st = new StringTokenizer(in.readLine());
  		}
  		return st.nextToken();
  	}
  
  	public int nextInt() throws Exception {
  		return Integer.parseInt(nextToken());
  	}
  
  	public int[] readArray(int n) throws Exception {
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextInt();
  		return a;
  	}
  
  	public long nextLong() throws Exception {
  		return Long.parseLong(nextToken());
  	}
  
  	public double nextDouble() throws Exception {
  		return Double.parseDouble(nextToken());
  	}
  
  }