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.
12
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
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
 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

2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 
 2 1 2 3 0 1 3 2 3 1 
1 2 3 4 
1 
1 1 1 0 1 1 1 1 0 0 0 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 

import java.io.*;
import java.util.*;
public class dhet {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		StringBuilder finalAnswer=new StringBuilder();
		finalAnswer.append(1).append('\n');
		StringBuilder sb=new StringBuilder();
		int t=sc.nextInt();
		while(t-->0) {
			int a=sc.nextInt();
			int b=sc.nextInt();
			int arr[]=new int[a];
			int brr[]=new int[a+1];
			for(int i=0;i<a;i++) {
				arr[i]=sc.nextInt();
				//System.out.println(arr[i]);
				//System.out.println(brr[arr[i]]);
				if(brr[arr[i]]>=b) {
					System.out.print(0+" ");
				}
				else {
					System.out.print(brr[arr[i]]+1+" ");
				}
				brr[arr[i]]++;
			}
			System.out.println();
		}
		//System.out.println();
		//System.out.println();
	}
	static int longestPalSubstr(String str)
	{
	    // get length of input String
	    int n = str.length();
	 
	    // All subStrings of length 1
	    // are palindromes
	    int maxLength = 1, start = 0;
	 
	    // Nested loop to mark start and end index
	    //for (int i = 0; i < str.length(); i++) {
	        for (int j = 0; j < str.length(); j++) {
	            int flag = 1;
	 
	            // Check palindrome
	            for (int k = 0; k < (j - 0 + 1) / 2; k++)
	                if (str.charAt(0 + k) != str.charAt(j - k))
	                    flag = 0;
	 
	            // Palindrome
	            if (flag!=0 && (j - 0 + 1) > maxLength) {
	                start = 0;
	                maxLength = j - 0 + 1;
	            }
	        }
	    //}
	    //printSubStr(str, start, start + maxLength - 1);
	 
	    // return length of LPS
	    return maxLength;
	}
//	static void printSubStr(String str, int low, int high)
//	{
//	    for (int i = low; i <= high; ++i)
//	        System.out.print(str.charAt(i));
//	}
	static int[] printDups(String str)
    {
        HashMap<Character, Integer> count = new HashMap<>();
        /*Store duplicates present
        in the passed string */
        for (int i = 0; i < str.length(); i++) {
            if (!count.containsKey(str.charAt(i)))
                count.put(str.charAt(i), 1);
            else
                count.put(str.charAt(i),
                          count.get(str.charAt(i)) + 1);
        }
        int how=0;
        int many=0;
        int arr[]=new int[2];
        /*Print duplicates in sorted order*/
        for (Map.Entry mapElement : count.entrySet()) {
            char key = (char)mapElement.getKey();
            int value = ((int)mapElement.getValue());
 
            if (value > 1) {
                how++;
            	many+=value;
            }
        }
        arr[0]=how;
        arr[1]=many;
        //System.out.println(many);
        return arr;
    }
}
