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.util.*;

public class abc{
    static final long mod=1_000_000_007;
    static long[] facts=new long[1_000_001], factInvs=new long[1_000_001];

    static void precomp() {
        facts[0]=1;
        for (int i=1; i<facts.length; i++) {
            facts[i]=mul(facts[i-1], i);
            factInvs[i]=inv(facts[i]);
        }
        factInvs[0]=1;
    }
    static long nCk(int n, int k) {
        if (k<0 || k>n) return 0;
        return mul(facts[n], mul(factInvs[k], factInvs[n-k]));
    }


    static final Random random=new Random();

    static void ruffleSort(int[] a) {
        int n=a.length;//shuffle, then sort
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long[] readArrayL(int n) {
            long[] a=new long[n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
    static long add(long a, long b) {
        return (a+b)%mod;
    }

    static long mul(long a, long b) {
        return (a*b)%mod;
    }

    static long fastPow(long base, long e) {
        if (e==0) return 1;
        long half=fastPow(base, e/2);
        if (e%2==0) return mul(half, half);
        return mul(half, mul(half, base));
    }

    static long inv(long x) {
        return fastPow(x, mod - 2);
    }
    static class pair{
     int x;int y;
     pair(){
         this.x=0;
         this.y=0;
     }
     pair(int x,int y) {
         this.x = x;
         this.y = y;
     }
    }
    public static void main(String[] args) throws IOException {
        FastScanner sc=new FastScanner();
        int t=sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt(), k = sc.nextInt();
            ArrayList<Integer> v=new ArrayList<>();
            int ans[]=new int[n];
            ArrayList<ArrayList<Integer>> idx=new ArrayList<>();
            for (int i = 0; i <n+5 ; i++) {
                idx.add(new ArrayList<>());
            }
            for (int i = 0; i <n ; i++) {
                int x=sc.nextInt();
                v.add(x);
                idx.get(x).add(i);
            }
            for (int i = 1; i <=n ; i++) {
                if(idx.get(i).size()>=k){
                    for (int j = 0; j <k ; j++) {
                        ans[(idx.get(i).get(j))]=j+1;
                    }
                    idx.get(i).clear();
                }
            }
            PriorityQueue<pair> pq=new PriorityQueue<>((c,d)->c.y-d.y);
            for (int i = 1; i <=k ; i++) {
                pq.add(new pair(0,i));
            }
            for (int i = 1; i <=n ; i++) {
                for (int j = 0; j <idx.get(i).size() ; j++) {
                    pair p=pq.poll();
                    ans[idx.get(i).get(j)]=p.y;
                    pq.add(new pair(p.x +1,p.y));
                }
            }
            int colors[]=new int[k+1];
            for (int i = 0; i <k ; i++) {
                colors[i]=0;
            }
            long m=mod;
            while(!pq.isEmpty()){
                colors[(pq.peek().y -1)]=pq.peek().x;
                m=Math.min(m,pq.peek().x);
                pq.poll();
            }
            for (int i = 0; i <n ; i++) {
                if(ans[i]!=0 && colors[ans[i]-1]>m){
                    colors[ans[i]-1]--;
                    ans[i]=0;
                }
            }
            for (int i:ans) {
                System.out.print(i+" ");
            }
            System.out.println();

        }
    }
}
