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.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t--!=0){
            Map<Integer,Stack<Integer>> mp=new HashMap<>();
            int n=sc.nextInt();
            int k=sc.nextInt();
            int[] tc=new int[n];
            int[] f=new int[n+5];
            int[] ans=new int[n];
            int[] kf=new int[k+1];
            int mx=n/k;
            for(int i=0;i<n;i++){
                tc[i]=sc.nextInt();
                f[tc[i]]++;
                if(mp.containsKey(tc[i])==true){
                    Stack<Integer> ta=mp.get(tc[i]);
                    ta.push(i);
                    mp.put(tc[i],ta);
                }else{
                    Stack<Integer> d=new Stack<>();
                    d.push(i);
                    mp.put(tc[i],d);
                }
                
            }
            int of=0;
            Stack<Integer> st=new Stack<>();
            Set<Integer> se=new HashSet<>();
            int am=0;
            for(int i=1;i<=n;i++){
                int c=tc[i-1];
                if(f[tc[i-1]]>=k ){
                    f[tc[i-1]]=-k;
                    kf[k]++;
                    ans[i-1]=-(f[tc[i-1]]);
                    //kf[-(f[tc[i-1]])]++;
                    f[tc[i-1]]++;
                    am++;
                }else if( f[tc[i-1]]<0){
                    ans[i-1]=-(f[tc[i-1]]);
                    kf[-(f[tc[i-1]])]++;
                    f[tc[i-1]]++;
                    
                }
                else if(f[c]<k && f[c]>0){
                    of++;
                    se.add(tc[i-1]);
                    
                    //st.push(i-1);
                }
                
            }
            mx=Math.min(am+(of/k),mx);
            int startC=1,color=0;
            for(int ele: se){
                int cn=0;
                Stack<Integer> idx=mp.get(ele);
                while(!idx.isEmpty() && cn<=k){
                    int id=idx.pop();
                    if(kf[color+1]<mx){
                        ans[id]=color+1;
                        kf[color+1]++;
                        color=(color+1)%k;
                        
                    }else cn++;
                    if(cn>k) break;
                }
                if(cn>k) break;
            }
            for(int i=0;i<n;i++){
                System.out.print(ans[i]+" ");
            }
            System.out.println();
        }
    }
    
}