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.*;


    public class Main{
        public static long mod = (long)1e9+7;
        public static ArrayList<Integer>[] graph;
        public static void main(String[] args) {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            InputReader in = new InputReader(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            int t = 1;
            t = in.nextInt();
            for(int i = 1;i<=t ;i++){

                solve(in,out);

            }
            out.close();
        }
        public static int max = Integer.MIN_VALUE;
        private static void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] list=  in.readIntArray(n);
            HashMap<Integer,Integer> hm = new HashMap<>();
            for(int u:list) {
                hm.put(u,hm.getOrDefault(u, 0)+1);
            }
            int cnt = 0;
            for(Integer k1:hm.keySet()) cnt += hm.get(k1)>k?k:hm.get(k1);
            int total = cnt/k;
            int[] res = new int[n];
            Pair[] plist = new Pair[n];
            for (int i = 0; i < plist.length; i++) {
                plist[i] = new Pair<Integer,Integer>(list[i],i);
            }
            Arrays.sort(plist);
            int color = 1;
            for (int i = 0; i < plist.length; i++) {
                if(i<k && total>0) {
                    res[(int)plist[i].second] = color;
                    if(color==k){
                        total--;
                        color = 1;
                    }
                    else color++;
                }
                else{
                    if(total<=0) break;
                    if((int)plist[i-k].first==(int)plist[i].first){
                        continue;
                    }
                    else{
                        res[(int)plist[i].second] = color;
                        if(color==k){
                        total--;
                        color = 1;
                        }
                        else color++;   
                    }

                } 
            }
            for(int o:res)out.print(o+" ");
            out.println();
        }
        
        
    }
    class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
    
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
    
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
    
        public int nextInt() {
            return Integer.parseInt(next());
        }
    
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public int[] readIntArray(int size){
            int[] list = new int[size];
            for(int i = 0;i<size;i++) list[i] = nextInt();
            return list;
        }
        public long[] readLongArray(int size){
            long[] list = new long[size];
            for(int i = 0;i<size;i++) list[i] = nextLong();
            return list;
        }
        public double[] readDoubleArray(int size){
            double[] list = new double[size];
            for(int i = 0;i<size;i++) list[i] = nextDouble();
            return list;
        }
        public String linereader() {
            String s = null;
            try{
                s = reader.readLine();
            }
            catch (IOException e) {
                throw new RuntimeException(e);
            }
            return s;
        }
    
    }
    class Pair<T extends Comparable<T>,K extends Comparable<K> > implements Comparable<Pair<T,K>>{
        T first;
        K second;
        public Pair(T first, K second){
            this.first  = first;
            this.second = second;
        }
        @Override
        public String toString(){
            return first+" "+second;
        }
        @Override
        public int compareTo(Pair<T,K> o) {
            // TODO Auto-generated method stub
            if((first).compareTo(o.first)!=0) return first.compareTo(o.first);
            else return second.compareTo(o.second);
        }

    }