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.lang.reflect.Array;
import java.math.BigInteger;
import java.security.cert.CollectionCertStoreParameters;
import java.util.*;



import static javax.swing.UIManager.get;
import static javax.swing.UIManager.getString;


public class Main {


    static class Pair implements Comparable<Pair>{
        int x=0,y=0;
        public Pair(int x1,int y1)
        { x=x1;
            y=y1;
        }


        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    static boolean checkPallindrome(int n)
    {    ArrayList<Integer> list = new ArrayList<>();
        while(n>0)
        { list.add(n%10);
            n/=10;
        }
        int low=0,high=list.size()-1;
        while(low<=high)
        {     if(list.get(low)!=list.get(high))
            return false;
            low++;
            high--;
        }
        return true;
    }


    static boolean check(String s)
    {

        int low=0,high=s.length()-1;
        while(low<=high)
        {     if(s.charAt(low)!=s.charAt(high))
            return false;
            low++;
            high--;
        }
        return true;
    }

         static  int min=Integer.MAX_VALUE;
    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
//(1)very very important**(never take the first problem for granted, always check the test cases) take 5 minutes more and check the edge cases
        // 5 minutes will not decreases rating as much as a wrong submission does it is easy u just think with an open mind and u will surely get the answer
        //(2)let ur brain consume the problem don't just jump to the solution. after reading the problem take a pause 1 minute
        //(3)go through the example test cases and also at least two of ur own test cases.Think of testcases which are difficult(edge cases).dry run ur concept

        int t = sc.nextInt();
        while (t-- > 0) {
            int n =sc.nextInt(),k=sc.nextInt();
            int[] a = sc.readArray(n);
            Map<Integer,Integer> map =new HashMap<>();
            ArrayList<ArrayList<Integer>> list = new ArrayList<>();

            for(int i=0;i<=n;i++)
                list.add(new ArrayList<Integer>());

            for(int i=0;i<n;i++)
            {  if(map.getOrDefault(a[i],0)<k)
            {
                list.get(a[i]).add(i);
                map.put(a[i],map.getOrDefault(a[i],0)+1);
            }
            }

            long count=0;
            for(ArrayList<Integer> x:list)
                if(x.size()<=k)
                    count+=x.size();
                else count+=k;

                long x = count/k;

//                System.out.println(x);
                 int count1=0;
                int[] b=  new int[n];
              Map<Integer,Integer> map1 = new HashMap<>();
//
//              for(int i=0;i<=n;i++)
//              { if(list.get(i).size()>0)
//              { System.out.println(i);
//                   for(int p:list.get(i))
//                         System.out.print(p+" ");
//                   System.out.println();
//              }
//
//              }
 long total=0;
 int flag=0;
              for(int i=0;i<=n;i++)
              {

                  if(list.get(i).size()>0)
              {     count1=1;

                  for(int p:list.get(i))
                      {      while(count1<=k&&map1.getOrDefault(count1,0)>=x)
                          count1++;
                              b[p] = count1;
                              total++;
                              map1.put(count1, map1.getOrDefault(count1, 0) + 1);
                              count1++;
                       if(total==x*k)
                       { flag=1;
                           break;
                       }
                         if(count1==k+1)
                         count1=1;
                      }
                  if(flag==1)
                      break;
              }

              }

              print(b);

        }
    }








//Things to check when u r getting wrong answer
// syntax of conditional operator y=(x==1)?1:0;
// 1-  check the flow of the code
//2- If ur stuck read the problem once again
//3- before submitting always check the output format of ur code
//4- don't check standings until problem B is done
//5- if u r thinking ur concept is correct but still u r getting wrong answer try to implement it in another way
//6- By default, java interpret all numeral literals as 32-bit integer values.
// If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.    example   long a = 600851475143L

//All the functions



    static void debug(String s)
    { System.out.println(s);

    }

    //collections.sort use merge sort instead of quick sort but arrays.sort use quicksort whose worst time complexity is O(n^2)
    static int[] sort(int[] a)
    {         ArrayList<Integer> list = new ArrayList<>();
        for(int i=0;i<a.length;i++)
            list.add(a[i]);

        Collections.sort(list);
        int ind=0;
        for(int x:list)
            a[ind++]=x;
        return a;

    }

    //function to print an array for debugging
    static void print(int[] a) {
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }

    static void printc(char[] a) {
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }


    //normal gcd function, always put the greater number as a and the smaller number as b
    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);

    }

    static int lcm(int a,int b)
    {
        return (a*b)/gcd(a,b);
    }


    //to find gcd and lcm for numbers of long data type
    static long gcdl(long a, long b) {
        if (b == 0)
            return a;
        return gcdl(b, a % b);

    }

    static long lcml(long a,long b)
    {
        return (a*b)/gcdl(a,b);
    }


    //Input Reader to read faster input
    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;
        }


        int[] readArray(int n)
        {  int[] a = new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();

            return a;
        }
    }



}

































































