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.math.*;
  import java.util.* ;
  import java.io.* ;
  @SuppressWarnings("unused")
  
  public class cf
  {	
  	
  	static final int mod = 1000000007 ;
  	static final double pi = 3.1415926536 ;
  	static boolean not_prime[] = new boolean[1000001] ;
  	static void sieve() {
  		for(int i=2 ; i*i<1000001 ; i++) {
  			if(not_prime[i]==false) {
  				for(int j=2*i ; j<1000001 ; j+=i) {
  					not_prime[j]=true ;
  				}
  			}
  		}not_prime[0]=true ; not_prime[1]=true ;
  	}	
  	
  	public static long bexp(long base , long power)
  	{
  		long res=1L ;
  		
  		base = base%mod ;
  		
  		while(power>0) {
  			if((power&1)==1) {
  				res=(res*base)%mod ;
  				power-- ;
  			}
  			else {
  				base=(base*base)%mod ;
  				power>>=1 ;
  			}
  		}
  		
  		return res ;
  	}
  
  	static long modInverse(long n, long p){return power(n, p - 2, p); }//calculates modulo inverse
  	static long power(long x, long y, long p)
  	{
  	  
  	    // Initialize result
  	    long res = 1;
  	    x = x % p;
  	  
  	    while (y > 0) {
  	  
  	        if (y % 2 == 1)
  	            res = (res * x) % p;
  	        
  	        x = (x * x) % p;
  	    }
  	  
  	    return res;
  	}
  	static long nCrModPFermat(int n, int r, long p)
  	{
  		if(n<r) return 0 ;
  	    if (r == 0)
  	        return 1;
  
  	    long[] fac = new long[n + 1];
  	    fac[0] = 1;
  	  
  	    for (int i = 1; i <= n; i++)
  	        fac[i] = fac[i - 1] * i % p;
  	  
  	    return (fac[n] * modInverse(fac[r], p) % p
  	            * modInverse(fac[n - r], p) % p)
  	        % p;
  	}
  	
  	private static long modular_add(long a, long b){ return ((a % mod) + (b % mod)) % mod; }
  	private static long modular_sub(long a, long b){ return ((a % mod) - (b % mod) + mod) % mod; }
  	private static long modular_mult(long a, long b){ return ((a % mod) * (b % mod)) % mod; }
  	
  	static long lcm(int a, int b){ return (a / gcd(a, b)) * b; }
  	private static long gcd(long a, long b){
  		if (b == 0) {
  			return a;
  		}
  		return gcd(b, a % b);
  	}
  	
  	static int cbits(int num) { return (int)(Math.log(num) /Math.log(2)) + 1; }
  
  	public static void main(String[] args)throws IOException//throws FileNotFoundException
      {
  		FastReader in = new FastReader() ;
  		StringBuilder op = new StringBuilder() ;
  		int T = in.nextInt() ;
  		for(int tt=0 ; tt<T ; tt++)
  		{
  			int n = in.nextInt(); 
  			int k =in.nextInt() ;
  			int a[] = in.readArray(n) ;
  			Map<Integer , ArrayList<Integer>> map = new HashMap<>() ;
  			for(int i : a)map.put(i, new ArrayList<>()) ;
  			for(int i=0 ; i<n ; i++) {
  				map.get(a[i]).add(i) ;
  			}
  			List<Integer> list = new ArrayList<>();
  			int res[] = new int[n];
  			//System.out.println(map);
  			for(int element : map.keySet())
  			{
  				if(map.get(element).size()>=k)
  				{
  					int index=0 ;
  					for(int color=1 ; color<=k ; color++)
  					{
  						res[map.get(element).get(index)] = color ;
  						index++ ;
  					}
  				}
  				else
  				{
  					for(int i : map.get(element))list.add(i) ;
  				}
  			}
  			//System.out.println(list);
  			int leave = list.size()%k ;
  			int col=1 ;
  			for(int i=0 ; i<list.size()-leave ; i++) {
  				res[list.get(i)]=col ; col+=1 ;
  				if(col>k)col=1 ;
  			}
  			for(int i : res)op.append(i+" ") ;
  			op.append("\n") ;
   		}
  		System.out.println(op.toString());
      }
  	
  	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;
          }
          
           int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
      }
  	
  	static void shuffleArray(int[] arr){
          int n = arr.length;
          Random rnd = new Random();
          for(int i=0; i<n; ++i){
              int tmp = arr[i];
              int randomPos = i + rnd.nextInt(n-i);
              arr[i] = arr[randomPos];
              arr[randomPos] = tmp;
          }   
  	}
  
  }