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: //Utilities
  import java.io.*;
  import java.util.*;
  
  public class C {
  	static int t;
  	static int n, k;
  	static int[] a, res;
  	static Queue<Integer>[] hold;
  	static Stack<Integer> took;
  	
  	public static void main(String[] args) throws IOException {
  		t = in.iscan();
  		while (t-- > 0) {
  			n = in.iscan(); k = in.iscan(); a = new int[n];
  			hold = new Queue[n+1];
  			for (int i = 0; i <= n; i++) {
  				hold[i] = new LinkedList();
  			}
  			for (int i = 0; i < n; i++) {
  				a[i] = in.iscan();
  				hold[a[i]].add(i);
  			}
  			took = new Stack<Integer>(); res = new int[n];
  			int idx = 1, cnt = 0;
  			for (int i = 1; i <= n; i++) {
  				int tmp = 0;
  				while (hold[i].size() > 0 && tmp < k) {
  					int cur = hold[i].poll();
  					res[cur] = idx;
  					took.push(cur);
  					if (idx != k) {
  						idx++;
  					}
  					else {
  						idx = 1;
  					}
  					cnt++; tmp++;
  				}
  			}
  			while (cnt % k != 0) {
  				res[took.pop()] = 0;
  				cnt--;
  			}
  			for (int i = 0; i < n; i++) {
  				out.print(res[i] + " ");
  			}
  			out.println();
  		}
  		out.close();
  	} 
  	
  	static INPUT in = new INPUT(System.in);
  	static PrintWriter out = new PrintWriter(System.out);
  	private static class INPUT {
  
  		private InputStream stream;
  		private byte[] buf = new byte[1024];
  		private int curChar, numChars;
  
  		public INPUT (InputStream stream) {
  			this.stream = stream;
  		}
  
  		public INPUT (String file) throws IOException {
  			this.stream = new FileInputStream (file);
  		}
  
  		public int cscan () throws IOException {
  			if (curChar >= numChars) {
  				curChar = 0;
  				numChars = stream.read (buf);
  			}
  			
  			if (numChars == -1)
  				return numChars;
  
  			return buf[curChar++];
  		}
  
  		public int iscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			int res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public String sscan () throws IOException {
  			int c = cscan ();
  			
  			while (space (c))
  				c = cscan ();
  
  			StringBuilder res = new StringBuilder ();
  
  			do {
  				res.appendCodePoint (c);
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res.toString ();
  		}
  
  		public double dscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			double res = 0;
  
  			while (!space (c) && c != '.') {
  				if (c == 'e' || c == 'E')
  					return res * UTILITIES.fast_pow (10, iscan ());
  				
  				res *= 10;
  				res += c - '0';
  				c = cscan ();
  			}
  
  			if (c == '.') {
  				c = cscan ();
  				double m = 1;
  
  				while (!space (c)) {
  					if (c == 'e' || c == 'E')
  						return res * UTILITIES.fast_pow (10, iscan ());
  
  					m /= 10;
  					res += (c - '0') * m;
  					c = cscan ();
  				}
  			}
  
  			return res * sgn;
  		}
  
  		public long lscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			long res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public boolean space (int c) {
  			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
  		}
  	}
  
  	public static class UTILITIES {
  
  		static final double EPS = 10e-6;
  
  		public static int lower_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] >= x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static int upper_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] > x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static long gcd (long a, long b) {
  			return b == 0 ? a : gcd (b, a % b);
  		}
  
  		public static long lcm (long a, long b) {
  			return a * b / gcd (a, b);
  		}
  
  		public static long fast_pow_mod (long b, long x, int mod) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  
  			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  		}
  
  		public static int fast_pow (int b, int x) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow (b * b, x / 2);
  
  			return b * fast_pow (b * b, x / 2);
  		}
  
  		public static long choose (long n, long k) {
  			k = Math.min (k, n - k);
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = val * (n - i) / (i + 1);
  
  			return val;
  		}
  
  		public static long permute (int n, int k) {
  			if (n < k) return 0;
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = (val * (n - i));
  
  			return val;
  		}
  		
  		// start of permutation and lower/upper bound template
  		public static void nextPermutation(int[] nums) {
  		    //find first decreasing digit
  		    int mark = -1;
  		    for (int i = nums.length - 1; i > 0; i--) {
  		        if (nums[i] > nums[i - 1]) {
  		            mark = i - 1;
  		            break;
  		        }
  		    }
  		 
  		    if (mark == -1) {
  		        reverse(nums, 0, nums.length - 1);
  		        return;
  		    }
  		 
  		    int idx = nums.length-1;
  		    for (int i = nums.length-1; i >= mark+1; i--) {
  		        if (nums[i] > nums[mark]) {
  		            idx = i;
  		            break;
  		        }
  		    }
  		 
  		    swap(nums, mark, idx);
  		 
  		    reverse(nums, mark + 1, nums.length - 1);
  		}
  		 
  		public static void swap(int[] nums, int i, int j) {
  		    int t = nums[i];
  		    nums[i] = nums[j];
  		    nums[j] = t;
  		}
  		 
  		public static void reverse(int[] nums, int i, int j) {
  		    while (i < j) {
  		        swap(nums, i, j);
  		        i++;
  		        j--;
  		    }
  		}
  		
  		static int lower_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] >= cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  	 
  		static int upper_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] > cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  		// end of permutation and lower/upper bound template
  	}
  }
  