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.util.*;

public class WonderfulColoring2
{
    private static boolean excellent(int[] array, int[] colors)
    {
        for (int i = 0; i < array.length - 1; i++)
        {
            for (int j = i + 1; j < array.length; j++)
            {
                if (array[i] == array[j] && colors[i] == colors[j])
                {
                    return false;
                }
            }
        }

        return true;
    }

    public static void main(String[] args)
    {
        Scanner scanner = new Scanner(System.in);
        int cases = scanner.nextInt();

        for (int i = 0; i < cases; i++)
        {
            int n = scanner.nextInt();
            int k = scanner.nextInt();

           int[] array = new int[n];
           // The number, and all the places it occurs.
           HashMap<Integer, ArrayList<Integer>> info = new HashMap<>();
           int[] res = new int[n];


           for (int j = 0; j < n; j++)
           {
               array[j] = scanner.nextInt();
               if (info.containsKey(array[j]))
               {
                   ArrayList<Integer> arrayList = info.get(array[j]);
                   arrayList.add(j);
                   info.replace(array[j], arrayList);
               }
               else
               {
                   ArrayList<Integer> newList = new ArrayList<>();
                   newList.add(j);
                   info.put(array[j], newList);
               }
           }

           //System.out.println("Let's do a test!");
           for (Map.Entry entry : info.entrySet())
           {
               ArrayList<Integer> arrayList = (ArrayList<Integer>) entry.getValue();
               if (arrayList.size() >= k)
               {
                   int curColor = 1;
                   for (Integer num : arrayList)
                   {
                       if (curColor > k)
                       {
                           res[num] = 0;
                       }

                       else
                       {
                           res[num] = curColor;
                           curColor++;
                       }

                   }
               }

               //System.out.println("The number " + entry.getKey() + " was found:");
               //System.out.println(entry.getValue().toString());
           }

           int curColor = 1;
           for (Map.Entry entry : info.entrySet())
           {
               ArrayList<Integer> arrayList = (ArrayList<Integer>) entry.getValue();
               if (arrayList.size() < k)
               {
                   for (Integer num : arrayList)
                   {
                       if (curColor > k)
                       {
                           curColor = 1;
                       }

                       res[num] = curColor;
                       curColor++;
                   }
               }
           }

           for (int j = 0; j < res.length; j++)
           {
               System.out.print(j < res.length - 1 ? res[j] + " " : res[j] + "\n");
           }
        }
    }
}
