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 

//package currentContest;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public class P3 {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
		int t;
		t = sc.nextInt();
		StringBuilder st = new StringBuilder();
		while (t-- != 0) {
			int n=sc.nextInt();
			int k=sc.nextInt();
			int a[]=new int[n];
			int c[]=new int[n];
			Arrays.fill(c, 0);
			for(int i=0;i<n;i++) {
				a[i]=sc.nextInt();
				c[a[i]-1]++;
			}
			int x=0;
			int co=0;
			for(int i=0;i<n;i++) {
				if(c[i]>=k) {
					x++;
				}else {
					co+=c[i];
				}
			}
			int per=x+co/k;
			if(per==0) {
				for(int i=0;i<n;i++) {
					st.append("0 ");
				}
				st.append("\n");
				continue;
			}
			ArrayList<ArrayDeque<Integer>> list=new ArrayList<>();
			for(int i=0;i<n;i++) {
				list.add(new ArrayDeque<>());
			}
			int curr=1;
			int token=0;
			for(int i=0;i<n;i++) {
				if(token>=per*k) {
					break;
				}
				for(int j=0;j<Math.min(k, c[i]);j++) {
					if(token>=per*k) {
						break;
					}
					list.get(i).add(curr);
					curr++;
					if(curr==k+1) {
						curr=1;
					}
					token++;
				}
			}
			for(int i=0;i<n;i++) {
				if(list.get(a[i]-1).size()==0) {
					st.append("0 ");
				}else {
					st.append(list.get(a[i]-1).getFirst()+" ");
					list.get(a[i]-1).removeFirst();
				}
			}
			st.append("\n");
		}
		System.out.println(st);
	}

	static FastReader sc = new FastReader();

	public static void solvegraph() {
		int n = sc.nextInt();

		int edge[][] = new int[n - 1][2];
		for (int i = 0; i < n - 1; i++) {
			edge[i][0] = sc.nextInt() - 1;
			edge[i][1] = sc.nextInt() - 1;
		}
		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			ad.add(new ArrayList<Integer>());
		}
		for (int i = 0; i < n - 1; i++) {
			ad.get(edge[i][0]).add(edge[i][1]);
			ad.get(edge[i][1]).add(edge[i][0]);
		}
		int parent[] = new int[n];
		Arrays.fill(parent, -1);
		parent[0] = n;
		ArrayDeque<Integer> queue = new ArrayDeque<>();
		queue.add(0);
		int child[] = new int[n];
		Arrays.fill(child, 0);
		ArrayList<Integer> lv = new ArrayList<Integer>();
		while (!queue.isEmpty()) {
			int toget = queue.getFirst();
			queue.removeFirst();
			child[toget] = ad.get(toget).size() - 1;
			for (int i = 0; i < ad.get(toget).size(); i++) {
				if (parent[ad.get(toget).get(i)] == -1) {
					parent[ad.get(toget).get(i)] = toget;
					queue.addLast(ad.get(toget).get(i));
				}
			}
			lv.add(toget);
		}

		child[0]++;
	}

	static void sort(int[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			int tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}

	static void sort(long[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			long tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}

	static String sort(String s) {
		Character ch[] = new Character[s.length()];
		for (int i = 0; i < s.length(); i++) {
			ch[i] = s.charAt(i);
		}
		Arrays.sort(ch);
		StringBuffer st = new StringBuffer("");
		for (int i = 0; i < s.length(); i++) {
			st.append(ch[i]);
		}
		return st.toString();
	}

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

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
