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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B{
    public static void main(String[] args){
        FS sc = new FS();
        PrintWriter pw = new PrintWriter(System.out);
        final int MAX = 1_00_000;
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int k = sc.nextInt();
            int[] a = sc.nextArray(n);
            if(k>n){
                StringBuilder sb = new StringBuilder();
                for(int i=0; i<n; i++){
                    sb.append((0)+" ");
                }
                System.out.println(sb.toString());
                continue;
            }
            int[] nums = new int[MAX+2];
            int max = Integer.MIN_VALUE;
            for(int i=0; i<n; i++){
                nums[a[i]]++;
                max = Math.max(a[i], max);
            }
            int[] colors = new int[n+1];
            int[] filled = new int[k];
            int last = 0;
            Arrays.fill(colors, -2);
            for(int i=0; i<n; i++){
                if(colors[i]!=-2) continue;
                // System.out.println(nums[a[i]]);
                if(nums[a[i]]>k){
                    int count = 0;
                    for(int j=i; j<n; j++){
                        if(a[i]==a[j] && count<k){ 
                            colors[j] = (last)%k;
                            filled[last%k]++;
                            last++;
                            last %= k;
                            count++;
                        }
                        else if(a[i]==a[j] && count>=k){ 
                            colors[j] = -1;
                            count++;
                        }
                    }
                }
                else{
                    for(int j=i; j<n; j++){
                        if(a[i]==a[j]){ 
                            colors[j] = (last)%k;
                            filled[last%k]++;
                            last++;
                            last %= k;
                        }
                    }
                }
            }
            int minfilled = Integer.MAX_VALUE;
            for(int fill : filled){
                minfilled = Math.min(fill, minfilled);
            }
            for(int i=0; i<k; i++){
                if(filled[i]>minfilled){
                    for(int j=0; j<n; j++){
                        if(filled[i]>minfilled && colors[j]==i){
                            colors[j] = -1;
                            filled[i]--;
                        }
                        else break;
                    }
                }
            }
            StringBuilder sb = new StringBuilder();
            for(int i=0; i<n; i++){
                sb.append((colors[i]+1)+" ");
            }
            System.out.println(sb.toString());
        }
        pw.flush();
        pw.close();
    }

    static class FS{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
    
        String next(){
            while(!st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored){
                }
            }
            return st.nextToken();
        }
        
        int[] nextArray(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[i] = nextInt();
            }
            return a;
        }
        
        long[] nextLongArray(int n){
            long[] a = new long[n];
            for(int i = 0; i < n; i++){
                a[i] = nextLong();
            }
            return a;
        }
        
        int nextInt(){
            return Integer.parseInt(next());
        }
        
        long nextLong(){
            return Long.parseLong(next());
        }
        
        double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}