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.PrintWriter;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class solution { 
    public static void main(String args[]) throws java.lang.Exception{ 
    	FastScanner s=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		int t=s.nextInt();
		for(int tt=0;tt<t;tt++) {
			int n=s.nextInt(),k=s.nextInt();
			int a[]=s.readArray(n);
			int ans[]=new int[n];
			HashMap<Integer,ArrayList<Integer>> map=new HashMap<>();
			for(int i=0;i<n;i++) {
				if(map.containsKey(a[i])) {
					ArrayList<Integer> d=map.get(a[i]);
					d.add(i);
					map.put(a[i],d);
				}else {
					ArrayList<Integer> d=new ArrayList<>();
					d.add(i);
					map.put(a[i],d);
				}
			}
			int c1=1;
			for (ArrayList<Integer> url : map.values()) {
				ArrayList<Integer> temp=url;
				for(int i=0;i<Math.min(k,temp.size());i++) {
					ans[temp.get(i)]=c1;
					c1++;
					if(c1==k+1) {
						c1=1;
					}
				}
			}
			if(c1!=1) {
				HashSet<Integer> set=new HashSet<Integer>();
				for(int i=1;i<c1;i++) {
					set.add(i);
				}
				for(int i=0;i<n;i++) {
					if(set.contains(ans[i])) {
						set.remove(ans[i]);
						ans[i]=0;
					}
				}
			}
			for(int i=0;i<n;i++) {
				out.print(ans[i]+" ");
			}
			out.println();
		}
		out.close();
    }
    public static int modulus(int a,int b) {
    	int ans=a%b;
    	if(ans<0) {
    		ans=b+ans;
    	}
    	return ans;
    }
    static void sort(long [] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
    static void sort(int [] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
    static int gcd(int a, int b){
      if (b == 0)
        return a;
      return gcd(b, a % b);
    }
    static void sortcol(int a[][],int c) {
		Arrays.sort(a, (x, y) -> {
			if(x[c]>y[c]) {
				return 1;
			}else {
				return -1;
			}
		});
	}
    public static void printb(boolean ans) {
    	if(ans) {
    		System.out.println("Yes");
    	}else {
    		System.out.println("No");
    	}
    }
    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
    static class Pair implements Comparable<Pair>{
    	int a , b; 
    	Pair(int x , int y){
    		a=x;
    		b=y;
    	}
    	public int compareTo(Pair o) {
    		return a != o.a ? a - o.a : b - o.b;
    	}
    }
}