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.*;
import java.io.*;

import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;
import static java.lang.Math.sqrt;

public class Solution {

	static Scanner scn = new Scanner(System.in);
	static StringBuilder sb = new StringBuilder();
	static PrintWriter out = new PrintWriter(System.out);

	// https://randommashupgenerator.netlify.app/
	// https://a2oj.com/Ladder6.html
	// https://codeforces.com/problemset?order=BY_RATING_ASC&tags=binary+search%2C1400-1400
	// https://www.webtoons.com/en/challenge/cp-series/binary-search/viewer?title_no=365151&episode_no=5

	public static void main(String[] HastaLaVistaLa) {
		int t = scn.nextInt();
		for(int tests = 0; tests < t; tests++) solve();
		out.println(sb);
		out.close();
	}

	public static void solve() {
		// Road To Specialist Day 53
		int n = scn.nextInt();
		int k = scn.nextInt();
		int[] a = new int[n];
		for(int i = 0; i < n; i++) a[i] = scn.nextInt();
		int[] sorted = new int[n];
		for(int i = 0; i < n; i++) sorted[i] = a[i];
		sort(sorted);
		int[] ans = new int[n];
		int[] freq = new int[n + 1];
		int cnt = 1;
		for(int i = 0; i < n; i++) {
			int ele = sorted[i];
			freq[ele]++;
			if(freq[ele] > k)
				continue;
			ans[i] = cnt;
			cnt++;
			cnt %= (k + 1);
			if(cnt == 0) cnt = 1;
		}
		int min = Integer.MAX_VALUE;
		int[] freqK = new int[k + 1];
		for(int i = 0; i < n; i++) freqK[ans[i]]++;
		for(int i : freqK) {
			if(i == 0) continue;
			min = min(min, i);
		}
		Arrays.fill(freq, 0);
		for(int i = 0; i < n; i++) {
			int ele = ans[i];
			freq[ele]++;
			if(freq[ele] > min) ans[i] = 0;
		}
		HashMap<Integer, List<Integer>> map = new HashMap<>();
		for(int i = 0; i < n; i++) {
			int sot = sorted[i];
			int answer = ans[i];
			if(map.containsKey(sot)) {
				List<Integer> l = map.get(sot);
				l.add(answer);
				map.put(sot, l);
			}else {
				List<Integer> l = new ArrayList<>();
				l.add(answer);
				map.put(sot, l);
			}
		}
		for(int i = 0; i < n; i++) {
			int ele = a[i];
			List<Integer> l = map.get(ele);
			sb.append(l.get(l.size() - 1) + " ");
			l.remove(l.size() - 1);
			map.put(ele, l);
		} 
		sb.append("\n");
	}
	public static void sort(int[] a) {
		List<Integer> l = new ArrayList<>();
		for(int i : a) l.add(i);
		Collections.sort(l);
		int s = 0;
		for(int i : l) a[s++] = i;
	}
}