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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class _1551_B2 {

    private static final FastScanner in = new FastScanner();
    private static final PrintWriter out = new PrintWriter(System.out);
    private static final PrintWriter err = new PrintWriter(System.err);

    public static void solve() {
        int n = in.nextInt();
        int k = in.nextInt();

        int[] arr = in.nextIntArray(n);
        HashMap<Integer, Integer> counts = new HashMap<>();
        for (int a : arr) {
            counts.put(a, Math.min(k, counts.getOrDefault(a, 0) + 1));
        }

        int lastUse = 0;
        PriorityQueue<Color> colors = new PriorityQueue<>();
        for (int i = 1; i <= k; i++) {
            colors.add(new Color(i, 0, lastUse++));
        }

        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            int el = arr[i];
            if (counts.get(el) > 0) {
                Color color = colors.poll();
                ans[i] = color.k;
                colors.add(new Color(color.k, color.count + 1, lastUse++));
                counts.put(el, counts.get(el) - 1);
            } else {
                ans[i] = 0;
            }
        }

        int base = -1;
        int[] cs = new int[k];
        while (!colors.isEmpty()) {
            Color c = colors.poll();
            if (base == -1) {
                base = c.count;
            }
            if (c.count > base) {
                cs[c.k - 1] = 1;
            }
        }

        for (int i = 0; i < n; i++) {
            if (ans[i] > 0 && cs[ans[i]-1] == 1) {
                cs[ans[i]-1] = 0;
                ans[i] = 0;
            }
        }

        for (int c : ans) {
            out.print(c + " ");
        }
        out.println();
    }

    static class Color implements Comparable<Color> {
        int k;
        int count;
        int lastUse;

        public Color(int k, int count, int lastUse) {
            this.k = k;
            this.count = count;
            this.lastUse = lastUse;
        }

        @Override
        public int compareTo(Color o) {
            return count == o.count ? lastUse - o.lastUse : count - o.count;
        }
    }

    // Boilerplate
    public static void main(String[] args) {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }

        err.close();
        out.close();
    }

    @SuppressWarnings("unused")
    static class FastScanner {

        StringTokenizer tok = new StringTokenizer("");
        BufferedReader in;

        FastScanner() {
            in = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (!tok.hasMoreElements()) {
                try {
                    tok = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }

            return tok.nextToken();
        }

        String nextLine() {
            String line;
            try {
                line = in.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

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

        int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        List<Integer> nextIntegerList(int n) {
            ArrayList<Integer> list = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                list.add(nextInt());
            }
            return list;
        }

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

        long[] nextLongArray(int n) {
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        List<Long> nextLongList(int n) {
            ArrayList<Long> list = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                list.add(nextLong());
            }
            return list;
        }

        ArrayList<Integer>[] nextGraph(int n, int m) {
            @SuppressWarnings("unchecked")
            ArrayList<Integer>[] adj = (ArrayList<Integer>[]) new ArrayList[n];

            for (int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }

            for (int i = 0; i < m; i++) {
                int v = nextInt();
                int t = nextInt();
                adj[v - 1].add(t - 1);
                adj[t - 1].add(v - 1);
            }

            return adj;
        }
    }
}