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.*;
import java.io.*;

public class Solution {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());

        BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));

        StringTokenizer st;

        while (t-- > 0) {

            // int n = Integer.parseInt(br.readLine());

            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int k = Integer.parseInt(st.nextToken());

            int arr[] = new int[n];
            int res[] = new int[n];

            HashMap<Integer, Integer> map = new HashMap<>();
            HashMap<Integer, ArrayList<Integer>> map2 = new HashMap<>();
            st = new StringTokenizer(br.readLine());

            int count = 0;
            for (int i = 0; i < n; i++) {
                int e = Integer.parseInt(st.nextToken());
                arr[i] = e;
                if (map.containsKey(e)) {
                    int val = map.get(e);
                    if (val == k)
                        continue;

                    else {
                        map.put(e, val + 1);
                        count++;
                        ArrayList<Integer> temp = map2.get(e);
                        temp.add(i);
                        map2.put(e,temp);
                    }
                }

                else {
                    map.put(e, 1);
                    ArrayList<Integer> temp = new ArrayList<>();
                    temp.add(i);
                    map2.put(e, temp);
                    count++;
                }
            }


            count = (count%k);

            Set<Integer> set = map.keySet();
            Iterator<Integer> iter = set.iterator();

            while(iter.hasNext())
            {
                int key = iter.next();
                int val = map.get(key);
                
                if(val>=count)
                {
                    val-=count;
                    map.put(key,val);
                    break;
                }

                else
                {
                    map.put(key,0);
                    count-=val;
                }
            }


            int c = 0;

            Set<Integer> set2 = map2.keySet();
            Iterator<Integer> iter2 = set2.iterator();

            
            while(iter2.hasNext())
            {
                int key = iter2.next();
                int corr = map.get(key);
                ArrayList<Integer> pos = map2.get(key);

                for(int i = 0; i < corr; i++)
                {
                    c++;
                    int color = c%k;
                    if(color == 0)
                    color = k;
                    res[pos.get(i)] = color;
                }
            }

            


            for(int i = 0; i < n; i++)
            {
                output.write(res[i] + " ");
            }

            output.write("\n");

            // output.write();
            // int n = Integer.parseInt(st.nextToken());

            // HashMap<Character, Integer> map = new HashMap<Character, Integer>();

            // if
            // output.write("YES\n");
            // else
            // output.write("NO\n");

            // long a = Long.parseLong(st.nextToken());
            // long b = Long.parseLong(st.nextToken());

            // if(flag == 1)
            // output.write("NO\n");
            // else
            // output.write("YES\n" + x + " " + y + " " + z + "\n");

            // output.write(n+ "\n");

        }

        output.flush();

    }

}