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.*;
import java.io.*;
public class Solution
{
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
    //static final long mod=(long)1e9+7;
    static final long mod=998244353L;
    public static long pow(long a,long p)
    {
        long res=1;
        while(p>0)
        {
            if(p%2==1)
            {
                p--;
                res*=a;
                res%=mod;
            }
            else
            {
                a*=a;
                a%=mod;
                p/=2;
            }
        }
        return res;
    }
    static class Pair
    {
        int u,v;
        Pair(int u,int v)
        {
            this.u=u;
            this.v=v;
        }
    }
    /*static class Pair implements Comparable<Pair>
    {
        int v,l;
        Pair(int v,int l)
        {
            this.v=v;
            this.l=l;
        }
        public int compareTo(Pair p)
        {
            return l-p.l; 
        }
    }*/
    static long gcd(long a,long b)
    {
        if(b%a==0)
        return a;
        return gcd(b%a,a);
    }
    public static class comp implements Comparator<Pair>
    {
        public int compare(Pair p1,Pair p2)
        {
            if(p1.v>p2.v)
            return 1;
            return -1;
        }
    }
    public static void upd1(int a[],int i)
    {
        for(;i<a.length;i+=(i&(-i)))
        a[i]+=1;
    }
    public static void upd2(long a[],int up)
    {
        for(int i=up;i<a.length;i+=(i&(-i)))
        a[i]=a[i]+1L*up;
    }
    public static int q1(int a[],int i)
    {
        int s=0;
        for(;i>0;i-=(i&(-i)))
        s+=a[i];
        return s;
    }
    public static long q2(long a[],int i)
    {
        long s=0;
        for(;i>0;i-=(i&(-i)))
        s+=a[i];
        return s;
    }
    public static void main(String args[])throws Exception
    {
        FastReader fs=new FastReader();
        PrintWriter pw=new PrintWriter(System.out);
        int tc=fs.nextInt();
        while(tc-->0)
        {
            int n=fs.nextInt();
            int k=fs.nextInt();
            int a[]=new int[n];
            for(int i=0;i<n;i++)
            a[i]=fs.nextInt();
            int cnt[]=new int[n+1];
            for(int i=0;i<n;i++)
            cnt[a[i]]++;
            int rem=n;
            TreeSet<Integer> ts[]=new TreeSet[n+1];
            for(int i=1;i<=n;i++)
            ts[i]=new TreeSet<>();
            for(int i=0;i<n;i++)
            ts[a[i]].add(i);
            int ans[]=new int[n];
            for(int i=1;i<=n;i++)
            {
                if(cnt[i]>=k)
                {
                    for(int j=1;j<=k;j++)
                    ans[ts[i].pollFirst()]=j;
                    rem-=cnt[i];
                }
            }
            int left=rem/k;
            ArrayList<Integer> list=new ArrayList<>();
            for(int i=0;i<n;i++)
            {
                if(cnt[a[i]]>=k)continue;
                list.add(i);
            }
            for(int j=0;j<left*k;j++)
            ans[list.get(j)]=(j%k)+1;
            for(int i:ans)
            pw.print(i+" ");
            pw.println();
        }
        pw.flush();
        pw.close();
    }
}