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.*;
import java.util.*;
import java.util.function.IntConsumer;


/*

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

1
3 1
1 2 3
 */
@SuppressWarnings("unchecked")

public class ws2 {
	
	static boolean usaco = false;
	static String file = "";
	static BufferedReader in;
	static PrintWriter out;
	public static void main(String[] args) throws Exception {
//		usaco = true;

		if(usaco) {
			in = new BufferedReader(new FileReader(file+".in"));
			out = new PrintWriter(new FileWriter(file+".out"));	
		}else {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new PrintStream(System.out));
		}
		int N = pi();
		for(int i = 0; i<N; i++) {
			int[] d = pl();
			int n = d[0];
			int k = d[1];
			int[][] nums = new int[n][3];
			int[] num = pl();
			for(int j = 0; j<num.length; j++) {
				nums[j][0] = num[j];
				nums[j][1] = j;
			}
			Arrays.sort(nums,(a,b)->a[0]-b[0]);
			int last = -1;
			int start = 0;
			nums[0][2] = 1;
			if(k==1)last = 0;
			int cur = 2<=k?2:1;
			for(int j = 1; j<nums.length; j++) {
				if(nums[j][0]==nums[j-1][0]) {
					if(j-start<k) {
						nums[j][2] = cur;
						cur++;
						
					}
				}else {
					start = j;
					nums[j][2] = cur;
					cur++;
				}

				if(cur>k) {
					last = j;
					cur = 1;
				}
			}
			for(int na = nums.length-1; na>last; na--) {
				nums[na][2] = 0;
			}
//			System.out.println(last+" "+Arrays.deepToString(nums));
			int[] res = new int[n];
			for(int[] ar: nums) {
				res[ar[1]] = ar[2];
			}
			for(int j =0; j<res.length; j++) {
				out.print(j==0?res[j]:" "+res[j]);
			}
			out.println();
		}
		in.close();
		out.close();
	}
	public static long[] pll() throws IOException {
		return pll(in.readLine());
	}
	public static long[] pll(String s) throws IOException {
		return Arrays.stream(s.split(" ")).mapToLong(Long::valueOf).toArray();
	}
	public static int[] pl() throws IOException {
		return pl(in.readLine());
	}
	public static int[] pl(String s) throws IOException {
		return Arrays.stream(s.split(" ")).mapToInt(Integer::valueOf).toArray();
	}
	public static double[] pld() throws IOException {
		return pld(in.readLine());
	}
	public static double[] pld(String s) throws IOException {
		return Arrays.stream(s.split(" ")).mapToDouble(Double::valueOf).toArray();
	}
	static void t(int n, IntConsumer consumer) {
	    for (int i = 0; i < n; i++) {
	        consumer.accept(i);
	    }
	}
	public static int pi() throws Exception {
		return Integer.valueOf(in.readLine());
	}
}
