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.math.*;
import java.util.* ;
import java.io.* ;
@SuppressWarnings("unused")

//Scanner s = new Scanner(new File("input.txt"));
//s.close();
//PrintWriter writer = new PrintWriter("output.txt");
//writer.close();
//[a-z] = [97-122] ------- [A-Z] = [65-90]

public class cf
{	
	
	static final int mod = 1000000007 ;
	static final double pi = 3.1415926536 ;
	static boolean not_prime[] = new boolean[1000001] ;
	static void sieve() {
		for(int i=2 ; i*i<1000001 ; i++) {
			if(not_prime[i]==false) {
				for(int j=2*i ; j<1000001 ; j+=i) {
					not_prime[j]=true ;
				}
			}
		}not_prime[0]=true ; not_prime[1]=true ;
	}	
	
	public static long bexp(long base , long power)
	{
		long res=1L ;
		
		base = base%mod ;
		
		while(power>0) {
			if((power&1)==1) {
				res=(res*base)%mod ;
				power-- ;
			}
			else {
				base=(base*base)%mod ;
				power>>=1 ;
			}
		}
		
		return res ;
	}

	static long modInverse(long n, long p){return power(n, p - 2, p); }//calculates modulo inverse
	static long power(long x, long y, long p)
	{
	  
	    // Initialize result
	    long res = 1;
	    x = x % p;
	  
	    while (y > 0) {
	  
	        if (y % 2 == 1)
	            res = (res * x) % p;
	        
	        x = (x * x) % p;
	    }
	  
	    return res;
	}
	static long nCrModPFermat(int n, int r, long p)
	{
		if(n<r) return 0 ;
	    if (r == 0)
	        return 1;

	    long[] fac = new long[n + 1];
	    fac[0] = 1;
	  
	    for (int i = 1; i <= n; i++)
	        fac[i] = fac[i - 1] * i % p;
	  
	    return (fac[n] * modInverse(fac[r], p) % p
	            * modInverse(fac[n - r], p) % p)
	        % p;
	}
	
	private static long modular_add(long a, long b){ return ((a % mod) + (b % mod)) % mod; }
	private static long modular_sub(long a, long b){ return ((a % mod) - (b % mod) + mod) % mod; }
	private static long modular_mult(long a, long b){ return ((a % mod) * (b % mod)) % mod; }
	
	static long lcm(int a, int b){ return (a / gcd(a, b)) * b; }
	private static long gcd(long a, long b){
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}
	
	static int cbits(int num) { return (int)(Math.log(num) /Math.log(2)) + 1; }

	public static void main(String[] args)throws IOException//throws FileNotFoundException
    {
		FastReader in = new FastReader() ;
		StringBuilder op = new StringBuilder() ;
		int T = in.nextInt() ;
		for(int tt=0 ; tt<T ; tt++)
		{
			int n = in.nextInt(); 
			int k =in.nextInt() ;
			int a[] = in.readArray(n) ;
			Map<Integer , ArrayList<Integer>> map = new HashMap<>() ;
			for(int i : a)map.put(i, new ArrayList<>()) ;
			for(int i=0 ; i<n ; i++) {
				map.get(a[i]).add(i) ;
			}
			List<Integer> list = new ArrayList<>();
			int res[] = new int[n];
			for(int element : map.keySet())
			{
				if(map.get(element).size()>=k)
				{
					int index=0 ;
					for(int color=1 ; color<=k ; color++)
					{
						res[map.get(element).get(index)] = color ;
						index++ ;
					}
				}
				else
				{
					for(int i : map.get(element))list.add(i) ;
				}
			}
			
			int leave = list.size()%k ;
			int col=1 ;
			for(int i=0 ; i<list.size()-leave ; i++) {
				res[list.get(i)]=col++ ;
			}
			for(int i : res)op.append(i+" ") ;
			op.append("\n") ;
 		}
		System.out.println(op.toString());
    }
	
	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;
		}
    }
	
	static void shuffleArray(int[] arr){
        int n = arr.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n-i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }   
	}

}