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 Self;

import java.util.Scanner;
import java.util.Set;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.Queue;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;

public class Main {

	static FastReader scn = new FastReader();

	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;
		}
	}

	static long mod = (long) (1e9 + 7);
	static int maxN = (int) 2e5 + 10;
	static StringBuilder sb = new StringBuilder();

	public static void main(String[] args) {

		int t = 1;
		t = scn.nextInt();

		while (t-- > 0) {

			int n = scn.nextInt();
			int k = scn.nextInt();

			int[] arr = input(n);

			int[] res = find(k, arr);

			for (int val : res)
				sb.append(val + " ");

			sb.append("\n");

		}

		System.out.println(sb);

	}

	private static class Pair implements Comparable<Pair> {
		int val;
		int freq;

		@Override
		public int compareTo(Pair o) {
			return o.freq - this.freq;
		}

		public Pair(int v, int f) {
			val = v;
			freq = f;
		}

	}

	private static int[] find(int k, int[] arr) {

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

		for (int val : arr)
			map.put(val, map.getOrDefault(val, 0) + 1);

		PriorityQueue<Pair> pq = new PriorityQueue<>();

		for (int key : map.keySet()) {
			pq.add(new Pair(key, map.get(key)));
		}

		HashMap<Integer, ArrayList<Integer>> ans = new HashMap<Integer, ArrayList<Integer>>();

		while (!pq.isEmpty()) {

			Pair top = pq.poll();
//			System.out.println(top.val + " " + top.freq);

			if (top.freq >= k) {
				ans.put(top.val, new ArrayList<>());
				for (int i = 1; i <= k; i++)
					ans.get(top.val).add(i);
			}

			else {
				pq.add(top);
				break;
			}
		}

		ArrayList<Integer> rem = new ArrayList<Integer>();

		while (!pq.isEmpty()) {

			Pair rp = pq.poll();

			while (rp.freq > 0) {
				rem.add(rp.val);
				rp.freq--;
			}
		}

//		System.out.println(rem);

		for (int i = 0; i < rem.size(); i += k) {

			int color = 1;

			for (int j = i; i + k <= rem.size() && j < i + k; j++) {
//				System.out.println("Hi");
				int val = rem.get(j);

				if (!ans.containsKey(val))
					ans.put(val, new ArrayList<>());

				ans.get(val).add(color);
				color++;
			}

		}

		int[] res = new int[arr.length];

		for (int i = 0; i < arr.length; i++) {
			int val = arr[i];
			ArrayList<Integer> temp = ans.get(val);

			if (temp != null && temp.size() > 0) {
				int size = temp.size();
				res[i] = temp.get(size - 1);
				temp.remove(size - 1);
			}
		}

		return res;

	}

	public static void yes() {
		sb.append("YES\n");
	}

	public static void no() {
		sb.append("NO\n");
	}

	public static void swap(int[] arr, int i, int j) {
		int temp = arr[i];
		arr[i] = arr[j];
		arr[j] = temp;
	}

	public static long[] longInput(int n) {

		long[] arr = new long[n];

		for (int i = 0; i < n; i++) {
			arr[i] = scn.nextLong();
		}

		return arr;

	}

	public static int[] input(int n) {

		int[] arr = new int[n];

		for (int i = 0; i < n; i++) {
			arr[i] = scn.nextInt();
		}

		return arr;

	}

}