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 Solution {

    public static long gcd(long a, long b){
        if(b==0){
            return a;
        }
        return gcd(b, a%b);
    }

    public static long gcdSum(long b){
        long a = 0;

        long temp = b;
        while(temp!=0){
            a = a + temp%10;
            temp = temp/10;
        }
        return gcd(a,b);

    }


    public  static class inputPair implements Comparable<inputPair>{
        private Integer a;
        private Integer b;

        public inputPair(int temp, int j) {
            a = temp;
            b = j;
        }
        public int compareTo(inputPair o1) {
            return this.a-o1.a;
        }

    }




    public static void main(String[] args){
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        //long  factArray[] = new long [42];


        for(int i =0;i<t;i++) {
            int n = s.nextInt();
            int k = s.nextInt();

            inputPair arr[] = new inputPair[n];
            int orig[] = new int[n];
            Set<inputPair> m[] = new HashSet[k+1];
            Set<Integer> dupm[] = new HashSet[k+1];

            for(int j =0;j<=k;j++){
                m[j] = new HashSet<>();
                dupm[j] = new HashSet<>();
            }

            for(int j =0;j<n;j++){
                int p = s.nextInt();
                arr[j] = new inputPair(p,j);
                //orig[j] = s.nextInt();
            }
            Arrays.sort(arr);
//            for(int j =0;j<n;j++){
//                System.out.println(arr[j].a + "==" + arr[j].b);
//            }

            int temp  = 0;
            //m[temp].add(arr[0]);
            //temp = (temp+1)%k;

            for(int j =0;j<n;j++){
                if(!dupm[temp].contains(arr[j].a)) {
                    m[temp].add(arr[j]);
                    dupm[temp].add(arr[j].a);
                    //System.out.println(arr[j].a+"->"+temp);
                    temp = (temp + 1) % k;
                }
                else {
                    m[k].add(arr[j]);
                    dupm[k].add(arr[j].a);
                    //System.out.println(arr[j].a+"-->"+k);
                }

            }

            int min=10000000;
            for(int j =0;j<k+1;j++){
                if(j<k&&m[j].size()<min)
                    min = m[j].size();
            }

            for(int j =0;j<k+1;j++){
                int sizeCurrent = 0;
                for(inputPair current : m[j]){
                    if(sizeCurrent>=min){
                        orig[current.b] = 0;
                        continue;
                    }
                    orig[current.b] = j+1;
                    sizeCurrent++;
                }
            }

            for(int j =0;j<n;j++){
                if(orig[j]==k+1)
                    orig[j] = 0;
                System.out.print(orig[j]+" ");
            }
            System.out.println();
        }


        return;
    }
}
