1551_B2. Wonderful Coloring - 2_35
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.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;

public class B2{
    public static class Pair implements Comparable<Pair> {
        int f, s;
    
        Pair(int f, int s) {
            this.f=f;
            this.s=s;
        }
    
        public int hashCode() {
            int hf = (int) (f ^ (f >>> 32));
            int hs = (int) (s ^ (s >>> 32));
            return 31 * hf + hs;
        }
    
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return s == other.s && f == other.f;
        }
    
        public int compareTo(Pair p) {
            if(this.f == p.f)
                return Long.compare(this.s, p.s);
            return Long.compare(this.f, p.f);
        }
    
        public String toString() {
            return "(" + f + ", " + s + ")";
        }
    }
    public static void main(String[] args){
        FS sc = new FS();
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int k = sc.nextInt();
            int[] a = sc.nextArray(n);
            ArrayList<Pair> list = new ArrayList<>();
            for(int i=0; i<n; i++){
                list.add(new Pair(a[i], i));
            }
            Collections.sort(list);
            // System.out.println(list.toString());
            int[] colors = new int[n];
            int[] filled = new int[k];
            colors[0] = 0;
            int prev = list.get(0).f;
            int count = 1, last = 1;
            filled[0]++;
            boolean unpainted = false;
            for(int i=1; i<n; i++){
                // int rem = n-i;
                // if(rem<k){
                //     unpainted = true;
                // }
                int curr = list.get(i).f;
                if(unpainted) colors[i] = -1;
                if((prev==curr && count>=k)){
                    colors[i] = -1;
                    count++;
                }
                else if(prev==curr && count<k){
                    count++;
                    colors[i] = (last)%k;
                    filled[last%k]++;
                    last++;
                    last %= k;
                }
                else if(prev!=curr||count<=k){
                    colors[i] = (last)%k;
                    filled[last%k]++;
                    last++;
                    last %= k;
                    count = 1;
                }
                prev = curr;
            }
            // System.out.println(Arrays.toString(colors));
            int[] ans = new int[n];
            for(int i=0; i<n; i++){
                ans[list.get(i).s] = colors[i];
            }
            // System.out.println(Arrays.toString(filled));
            int minfilled = Integer.MAX_VALUE;
            for(int fill : filled){
                minfilled = Math.min(fill, minfilled);
            }
            // System.out.println(Arrays.toString(filled)+" "+minfilled);
            for(int i=0; i<k; i++){
                if(filled[i]>minfilled){
                    for(int j=0; j<n; j++){
                        // System.out.println(i+" "+j+" "+Arrays.toString(filled));
                        if(filled[i]>minfilled && ans[j]==i){
                            ans[j] = -1;
                            filled[i]--;
                        }
                        else if(filled[i]<=minfilled) break;
                    }
                }
            }
            StringBuilder sb = new StringBuilder();
            for(int i=0; i<n; i++){
                sb.append((ans[i]+1)+" ");
            }
            System.out.println(sb);
        }
        pw.flush();
        pw.close();
    }

    static class FS{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
    
        String next(){
            while(!st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored){
                }
            }
            return st.nextToken();
        }
        
        int[] nextArray(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[i] = nextInt();
            }
            return a;
        }
        
        long[] nextLongArray(int n){
            long[] a = new long[n];
            for(int i = 0; i < n; i++){
                a[i] = nextLong();
            }
            return a;
        }
        
        int nextInt(){
            return Integer.parseInt(next());
        }
        
        long nextLong(){
            return Long.parseLong(next());
        }
        
        double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}