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 codeforce.div3.r734;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.ArrayList;
import java.util.List;
import java.util.HashMap;
import java.util.Map;
import java.util.Set;
import java.util.HashSet;

import static java.lang.System.in;
import static java.lang.System.out;
import static java.util.stream.Collectors.joining;


/**
 * @author pribic (Priyank Doshi)
 * @see <a href="https://m1.codeforces.com/contest/1551/problem/B2" target="_top">https://m1.codeforces.com/contest/1551/problem/B2</a>
 * @since 23/07/21 8:26 PM
 */
public class B2 {
  static FastScanner sc = new FastScanner(System.in);

  public static void main(String[] args) {
    try (PrintWriter out = new PrintWriter(System.out)) {
      int T = sc.nextInt();
      for (int tt = 1; tt <= T; tt++) {
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] arr = new int[n];
        Map<Integer, List<Integer>> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
          arr[i] = sc.nextInt();
          map.putIfAbsent(arr[i], new ArrayList<>());
          if (map.get(arr[i]).size() < k)
            map.get(arr[i]).add(i);
        }
        List<Integer> indexes = map.values().stream().flatMap(Collection::stream).collect(Collectors.toList());
        while (indexes.size() % k != 0) indexes.remove(indexes.size() - 1);
        indexes.sort((i, j) -> arr[i] - arr[j]);
        int[] ans = new int[n];
        for (int i = 0; i < indexes.size(); i++) {
          ans[indexes.get(i)] = i % k + 1;
        }

        StringBuilder str = new StringBuilder();
        for (int num : ans) str.append(num).append(" ");
        System.out.println(str);
      }
    }
  }

  static class FastScanner {
    BufferedReader br;
    StringTokenizer st;

    public FastScanner(File f) {
      try {
        br = new BufferedReader(new FileReader(f));
      } catch (FileNotFoundException e) {
        e.printStackTrace();
      }
    }

    public FastScanner(InputStream f) {
      br = new BufferedReader(new InputStreamReader(f), 32768);
    }

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

    boolean hasMoreTokens() {
      while (st == null || !st.hasMoreTokens()) {
        String s = null;
        try {
          s = br.readLine();
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (s == null)
          return false;
        st = new StringTokenizer(s);
      }
      return true;
    }

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

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

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